Integral Calculus - Exercises

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1 Integral Calculus - Eercises 6. Antidifferentiation. The Indefinite Integral In problems through 7, find the indicated integral.. Solution. = = + C = + C.. e Solution. e =. ( 5 +) Solution. ( 5 +) = e =e + C. 5 + = = C = = C. 4. ³ + Solution. µ + = + = = ln ( ) + + C = = ln C.

2 INTEGAL CALCULUS - EXECISES 4 5. e + 6 +ln Solution. µe + 6 +ln = e +6 +ln = e +6ln +(ln) + C. = 6. + Solution. + = + = = C = = C = = C. 7. ( ) 5 Solution. µ ( ) 5 = = ( 5 + ) = ( 5 + ) = = C = = C. 8. Find the function f whose tangent has slope +for each value of and whose graph passes through the point (, ). Solution. The slope of the tangent is the derivative of f. Thus f () = + and so f() is the indefinite integral f() = f () = µ + = = C.

3 INTEGAL CALCULUS - EXECISES 4 Using the fact that the graph of f passes through the point (, ) you get = 4 +++C or C = 5 4. Therefore, the desired function is f() = It is estimated that t years from now the population of a certain lakeside community will be changing at the rate of.6t +.t +.5 thousand people per year. Environmentalists have found that the level of pollution in the lake increases at the rate of approimately 5 units per people. By how much will the pollution in the lake increase during the net years? Solution. Let P (t) denote the population of the community t years from now. Then the rate of change of the population with respect to time is the derivative dp dt = P (t) =.6t +.t +.5. It follows that the population function P (t) is an antiderivative of.6t +.t +.5. Thatis, P (t) = P (t)dt = (.6t +.t +.5)dt = =.t +.t +.5t + C for some constant C. During the net years, the population will grow on behalf of P () P () = C C = =.6+.4+=thousand people. Hence, the pollution in the lake will increase on behalf of 5 =5 units.. Anobjectismovingsothatitsspeedaftert minutes is v(t) =+4t+t meters per minute. How far does the object travel during rd minute? Solution. Let s(t) denote the displacement of the car after t minutes. Since v(t) = ds = dt s (t) it follows that s(t) = v(t)dt = ( + 4t +t )dt = t +t + t + C. During the rd minute, the object travels s() s() = C 4 8 C = = meters.

4 INTEGAL CALCULUS - EXECISES 4 Homework In problems through, find the indicated integral. Check your answers by differentiation ( +6) e + ³ 8. + ³ 9. + e ( ). ( +) 4. Find the function whose tangent has slope 4 +for each value of and whose graph passes through the point (, ). 5. Find the function whose tangent has slope +6 for each value of and whose graph passes through the point (, 6). 6. Find a function whose graph has a relative minimum when =and a relative maimum when =4. 7. It is estimated that t months from now the population of a certain town will be changing at the rate of 4+5t people per month. If the current population is, what will the population be 8 months from now? 8. An environmental study of a certain community suggests that t years from now the level of carbon monoide in the air will be changing at therateof.t +. parts per million per year. If the current level of carbon monoide in the air is.4 parts per million, what will the level be years from now? 9. After its brakes are applied, a certain car decelerates at the constant rate of 6 meters per second per second. If the car is traveling at 8 kilometers per hour when the brakes are applied, how far does it travel before coming to a complete stop? (Note: 8 kmph is the same as mps.). Suppose a certain car supplies a constant deceleration of A meters per second per second. If it is traveling at 9 kilometersperhour(5 meters per second) when the brakes are applied, its stopping distance is 5 meters. (a) What is A?

5 INTEGAL CALCULUS - EXECISES 44 (b) What would the stopping distance have been if the car had been traveling at only 54 kilometers per hour when the brakes were applied? (c) At what speed is the car traveling when the brakes are applied if the stopping distance is 56 meters? esults C C 7. + C C C ln + C 5 7. e C p ( ) + + C 9. ln + + e + + C. +ln + C C C C 4. f() = + 5. f() = f() = 5 +4; not unique parts per million meters. (a) A =6.5 (b) 4 meters (c).7 kilometers per hour

6 INTEGAL CALCULUS - EXECISES Integration by Substitution In problems through 8, find the indicated integral.. ( +6) 5 Solution. Substituting u = +6and du =, youget ( +6) 5 = u 5 du = u6 + C = ( +6)6 + C.. [( ) 5 +( ) +5] Solution. Substituting u = and du =, youget ( ) 5 +( ) +5 = (u 5 +u +5)du = = 6 u6 + u +5u + C = = 6 ( )6 +( ) +5( ) + C. Since, for a constant C, C 5 is again a constant, you can write ( ) 5 +( ) +5 = 6 ( )6 +( ) +5 + C.. e Solution. Substituting u = and du =, youget e = e u du = eu + C = e + C e 6 Solution. Substituting u = 6 and du = 6 5, youget 5 e 6 = e u du = 6 6 eu + C = 6 e 6 + C Solution. Substituting u = 5 +and 5 du =4, you get = 5 u du = 5 ln u + C = 5 ln C.

7 INTEGAL CALCULUS - EXECISES Solution. Substituting u = 4 +6 and 5 du =( 5), you get = 5 du = 5 u = C. u du = 5 u + C = 7. ln Solution. Substituting u =ln and du =, youget ln = du =ln u + C =ln ln + C. u 8. ln Solution. Substituting u =ln and du =, youget ln ln = = udu = u + C =(ln) + C. 9. Use an appropriate change of variables to find the integral ( +)( ) 9. Solution. Substituting u =, u += +and du =, you get ( +)( ) 9 = (u +)u 9 du = (u +u 9 )du = = u + u + C = = ( ) + ( ) + C.. Use an appropriate change of variables to find the integral ( +). Solution. Substituting u =, u +4= + and du =, you

8 INTEGAL CALCULUS - EXECISES 47 get ( +) = (u +4) udu = u du + u du = = 5 u 5 + u + C = 5 u u + C = = 5 ( ) ( ) + C = Homework = 5 ( ) + 4 ( ) +C = = ( ) µ C = = µ ( ) +C. In problems through 8, find the indicated integral and check your answer by differentiation.. e e 5. e 6. ( +) ( +) ( +5) ( +)( + +5). ( )e ( +6) ln 5 6. (ln ) 7. ln( +) 8. e + In problems 9 through, use an appropriate change of variables to find the indicated integral ( 5) 6 ( 4) Find the function whose tangent has slope +5for each value of and whose graph passes through the point (, ). 5. Find the function whose tangent has slope whose graph passes through the point (, 5). for each value of and

9 INTEGAL CALCULUS - EXECISES A tree has been transplanted and after years is growing at the rate of + meters per year. After two years it has reached a height (+) of five meters. How tall was it when it was transplanted? 7. It is projected that t years from now the population of a certain country will be changing at the rate of e.t million per year. If the current population is 5 million, what will the population be years from now? esults.. 5 e5 + C. (4 ) 4 +C 6. ln +5 + C 4. e + C 5. e + C 6. ( +) 6 + C 7. ( +8) 4 +8+C 8. ( +) C 9. ( +5) + C. 6 ( + +5) + C. e + C. 5 ln C. ( +6) + C 4. ln + C 5. ln 5 + C 6. ln + C 7. ln ( +)+C 8. e + C 9. +ln + C. 5 ( +) + ( +) ++C. + C. 7 +ln 4 + C ( 5) 5 4( 5) ln + + C f() = ( +5) f() = ln meters 7. 6 million

10 INTEGAL CALCULUS - EXECISES Integration by Parts In problems through 9, use integration by parts to find the given integral.. e. Solution. Since the factor e. is easy to integrate and the factor is simplified by differentiation, try integration by parts with g() =e. and f() =. Then, G() = e. =e. and f () = and so e. = e. e. =e. e. + C = = ( )e. + C.. ( )e Solution. Since the factor e is easy to integrate and the factor is simplified by differentiation, try integration by parts with g() =e and f() =. Then, G() = e = e and f () = and so ( )e = ( )( e ) e = = ( )e +e + C =( )e + C.. ln Solution. In this case, the factor is easy to integrate, while the factor ln is simplified by differentiation. This suggests that you try integration by parts with g() = and f() =ln. Then, G() = = and f () = =

11 INTEGAL CALCULUS - EXECISES 5 and so ln = ln = ln = = ln + C = ln + C. 4. Solution. Since the factor is easy to integrate and the factor is simplified by differentiation, try integration by parts with g() = and f() =. Then, G() = = ( ) and f () = and so = ( ) + = ( ) + ( ) = µ 5 ( ) 5 + C = = ( ) 4 5 ( ) 5 + C = = ( ) 4 5 ( ) + C. 5. ( +)( +) 6 Solution. Since the factor ( +) 6 is easy to integrate and the factor +is simplified by differentiation, try integration by parts with g() =( +) 6 and f() = +. Then, G() = ( +) 6 = 7 ( +)7 and f () = and so ( +)( +) 6 = 7 ( +)( +)7 ( +) 7 = 7 = 7 ( +)( +)7 7 8 ( +)8 + C = = 56 [8( +) ( +)]( +)7 + C = = 56 (7 +6)( +)7 + C.

12 INTEGAL CALCULUS - EXECISES 5 6. e Solution. Since the factor e is easy to integrate and the factor is simplified by differentiation, try integration by parts with g() =e and f() =. Then, G() = e = e and f () = and so e = e e. To find e, you have to integrate by parts again, but this time with g() =e and f() =. Then, G() = e and f () = and so e = e e. To find e, you have to integrate by parts once again, this time with g() =e and f() =. Then, G() = e and f () = and so e = e e = e 4 e. Finally, e = e = µ e µ e 4 e + C = e + C.

13 INTEGAL CALCULUS - EXECISES 5 7. ln Solution. In this case, the factor is easy to integrate, while the factor ln is simplified by differentiation. This suggests that you try integration by parts with Then, and so ln G() = g() = and f() =ln. = = and f () = = ln + = ln 4 + C. = ln + µ + C = 8. e ³ Solution. First rewrite the integrand as e, and then integrate by parts with g() =e and f() =. Then, from Eercise 6.. you get G() = e = and f () = e and so e = e e = e e + C = = ( )e + C. 9. ( ) Solution. First rewrite the integrand as [( ) ],andthen integrate by parts with g() =( ) and f() =. Then G() = ( ) and f () =.

14 INTEGAL CALCULUS - EXECISES 5 Substituting u = and du =, youget G() = ( ) = u du = u = ( ). Then ( ) = ( ) ( ) = = ( ) ( ) + C = = ( ) 64 ( ) + C. (a) Use integration by parts to derive the formula n e a = a n e a n n e a. a (b) Use the formula in part (a) to find e 5. Solution. (a) Since the factor e a is easy to integrate and the factor n is simplified by differentiation, try integration by parts with g() =e a and f() = n. Then, G() = e a = a ea and f () =n n and so n e a = a n e a n a n e a. (b) Apply the formula in part (a) with a =5and n =to get e 5 = 5 e 5 e 5. 5 Again, apply the formula in part (a) with a =5and n =to find the new integral e 5 = 5 e 5 e 5. 5

15 INTEGAL CALCULUS - EXECISES 54 Homework Once again, apply the formula in part (a) with a =5and n = to get e 5 = 5 e5 e 5 = 5 5 e5 5 e5 and so e 5 = 5 e 5 5 = 5 5 e 5 µ 5 µ e5 5 e5 + C = e 5 + C. In problems through 6, use integration by parts to find the given integral.. e. e. e 5 4. ( )e 5. ln ( +) e. e. e. ln 4. (ln ) 5. ln 6. 7 ( 4 +5) 8 7. Find the function whose tangent has slope ( +)e for each value of and whose graph passes through the point (, 5). 8. Find the function whose tangent has slope ln for each value of > and whose graph passes through the point (, ). 9. After t seconds, an object is moving at the speed of te t meters per second. Epress the distance the object travels as a function of time.. It is projected that t years from now the population of a certain city will be changing at the rate of t ln t +thousand people per year. If the current population is million, what will the population be 5 years from now?

16 INTEGAL CALCULUS - EXECISES 55 esults.. ( +)e + C. ( 4)e + C. 5( +5)e 5 + C 4. ( )e + C 5. (ln )+C C )9 9 +) + C 8. ( +) 4 +) + C 9. ( +) +) + C. ( + +)e + C. + 9 e + C. ( +6 6)e + C. ln 9 + C 4. ln ln + + C 5. (ln +)+C ( 4 +5) 9 6 (4 +5) + C 7. f() = ( +)e + e f() = 4 ln 5 ln 9.. s(t) = (t +)e t

17 INTEGAL CALCULUS - EXECISES The use of Integral tables In Problems through 5, use one of the integration formulas from a table of integrals (see Appendi) to find the given integral.. + Solution. First rewrite the integrand as + = p ( +) 4 andthensubstituteu = +and du = to get + = p ( +) 4 = du u 4 = = ln u + u 4 + C =ln ++ p ( +) 4 + C = = ln C.. 6 Solution. First, rewrite the integrand as 6 = ( + ) = 4 ( +) = 4 ( +) andthensubstituteu = +and du = to get = 6 4 ( = du +) 4 = u = 8 ln 4 + u 4 u + C = 8 ln 4+u 4 u + C = = 8 ln 7+ + C.. ( +) Solution. First rewrite the integrand as ( +) =( +) += Aplly appropriate formulas (see Appendi, formulas 9 and ), to get + = 8 ( + ) + 8 ln C

18 INTEGAL CALCULUS - EXECISES 57 and + = ++ ln C. Combine these results, to conclude that ( +) = 8 ( +5) p ( +)+ 8 ln C. 4. e Solution.Aplly appropriate formula (see Appendi, formula ), to get = e + = e ln + e + C = = + ln e + C = + ln e + ln + C. Since the epression ln is a constant, you can write e = + ln e + C. 5. (ln ) Solution. Aplly the reduction formula (see Appendi, formula 9) (ln ) n = (ln ) n n (ln ) n to get (ln ) = (ln ) (ln ) = µ = (ln ) (ln ) = (ln ) (ln ) +6 (ln ) µ ln = = = (ln ) (ln ) +6 ln 6 + C. Homework In Problems through, use one of the integration formulas listed in this section to find the given integral... ( ) 4( 5) e. e

19 INTEGAL CALCULUS - EXECISES 58 Locate a table of integrals and use it to find the integrals in Problems through (ln ) One table of integrals lists the formula p ± p =ln + p ± p p while another table lists p ± p =ln p + ± p. Can you reconcile this apparent contradiction? 8. The following two formulas appear in a table of integrals: p = p ln p + p and a + b = ab ln a + ab a ab (for ab ). (a) Use the second formula to derive the first. (b) Apply both formulas to the integral. Which do you find 9 4 easier to use in this problem? esults.. ln + C. ln 5 + C. ln C 4. ln + q C 5. ln C 6. ln + + C 7. ln + + C 8. 4ln + C e + C. ( )e + C. ln + C. ( +4) +4+ p ( +4)+C 6. (ln ) ln + + C 4. ln C 5. p q (4 )+C 6. ln C 7. ln a b =lna ln b 8. ln + + C

20 INTEGAL CALCULUS - EXECISES The Definite Integral In problems through 7 evaluate the given definite integral. (e t e t ) dt ln Solution. e t e t dt = e t + e t ln ln = e + e e ln e ln = = e + e e ln e ln = e + e = = e + e 5.. ( +6)4 Solution. Substitute u = +6. Then du =, u( ) =, and u() = 6. Hence,. ( +) ( +6) 4 = 6 u 4 du = u5 6 = 65 = Solution. Substitute u = +. Then du =, u() =, and u() = 9. Hence, 4. e e ln ( +) = 9 u du = u 9 = = Solution. Substitute u =ln. Thendu =, u(e) =,andu(e )=. Hence, e e ln = du u =ln u =ln ln = ln. 5. e t ln tdt Solution. Since the factor t is easy to integrate and the factor ln t is simplified by differentiation, try integration by parts with g(t) =t and f(t) =lnt

21 INTEGAL CALCULUS - EXECISES 6 Then, G(t) = tdt = t and f (t) = t and so e t ln tdt = t ln t e e tdt = t ln t 4 t = 8 e ln e 6 e 8 ln + 6 = 8 e 6 e + 6 = = 6 e + 6 = 6 (e +). 6. e Solution. Apply the reduction formula n e a = a n e a n a twice to get e = e e = e = n e a = e e + e = µ = e e + µ 4 e = + e 4 µ = + e 4 4 e = 4 e 4 = 4 (e ) t te dt Solution. Integrate by parts with g(t) =e 5 t and f(t) =t = Then, G(t) = e 5 t dt =e 5 t and f (t) = and so 5 te 5 t dt = te 5 t 5 5 = (t )e 5 t ³ = + 4e 4 = 5 ³ 5 e 5 t dt = te 5 t 4e 5 t = = ( 5)e ( )e 4 = 4e 4.

22 INTEGAL CALCULUS - EXECISES 6 (a) Show that b a f() + c b f() = c a f(). (b) Use the formula in part (a) to evaluate. (c) Evaluate 4 ( + ). Solution. (a) By the Newton-Leibniz formula, you have b a f() + c b f() = F (b) F (a)+f (c) F (b) = = F (c) F (a) = c a f(). (b) Since = for and = for, youhaveto break the given integral into two integrals = ( ) = =+ = and = Thus, = = + = =. = + =. (c) Since = + for and = for, you get 4 ( + ) = = [ + ( +)] + ( +4) [ + ( )] = ( ) = = ( +4) + ( ) = = (a) Show that if F is an antiderivative of f, then b a f( ) = F ( b)+f ( a) 4 =

23 INTEGAL CALCULUS - EXECISES 6 (b) A function f is said to be even if f( ) =f(). [For eample, f() = is even.] Use problem 8 and part (a) to show that if f is even, then a a f() = f() a (c) Use part (b) to evaluate and. (d) A function f is said to be odd if f( ) = f(). Useproblem8 and part (a) to show that if f is odd, then a a f() =. (e) Evaluate. Solution. (a) Substitute u =. Thendu =, u(a) = a and u(b) = b. Hence, b a f( ) = b a (b) Since f( ) =f(), you can write a a f() = By the part (a), you have Hence, a f(u)du = F (u) b a = F ( b)+f ( a). a f( ) + a f(). f( ) = F () + F ( ( a)) = F (a) F () = = a a a f(). f() = a f(). (c) Since f() = is an even function, you have Analogously, = = = = = =. = = 6.

24 INTEGAL CALCULUS - EXECISES 6 (d) Since f( ) = f(), you can write a a f() = a f( ) + a f() = = F () F (a)+f (a) F () =. (e) Since f() = is an odd function, you have =.. It is estimated that t days from now a farmer s crop will be increasing at the rate of.t +.6t +bushels per day. By how much will the value of the crop increase during the net 5 days if the market price remains fied at euros per bushel? Solution. Let Q(t) denote the farmer s crop t days from now. Then the rate of change of the crop with respect to time is dq dt =.t +.6t +, and the amount by which the crop will increase during the net 5 days is the definite integral 5 Q(5) Q() =.t +.6t + =.t +.t + t 5 = = =5bushels. Hence, the value of the market price will increase by 5 =75euros. Homework In problems through 7, evaluate the given definite integral. (4 +). (5 + ) 5. ( + t ³ 9 t +t ) dt 4. dt t t+dt t 6 7. ( ) 8. ( 4) t+ dt. (t + t) t 4 +t +dt 6.. e+ +. (t +)(t e )9 dt 4. ln tdt e 5. e (ln ) ( + t)e.t dt

25 INTEGAL CALCULUS - EXECISES A study indicates that months from now the population of a certain town will be increasing at the rate of 5+ people per month. By how much will the population of the town increase over the net 8 months? 9. It is estimated that the demand for oil is increasing eponentially at the rate of percent per year. If the demand for oil is currently billion barrels per year, how much oil will be consumed during the net years?.anobjectismovingsothatitsspeedaftert minutes is 5+t +t meters per minute. How far does the object travel during the nd minute? esults ln ln. e e + 5. e e people billion barrels. 5 meters

26 INTEGAL CALCULUS - EXECISES Area and Integration In problems through 9 find the area of the region.. is the triangle with vertices ( 4, ), (, ) and (, 6). Solution. From the corresponding graph (Figure 6.) you see that the region in question is bellow the line y = +4above the ais, and etends from = 4 to =. y 6 4 y= Figure 6.. Hence, A = 4 ( +4) = µ +4 4 =(+8) (8 6) = 8.. is the region bounded by the curve y = e, the lines =and =ln,andthe ais. Solution. Since ln =ln ln = ln '.7, from the correspondinggraph(figure6.)youseethattheregioninquestionis bellow the line y = e above the ais, and etends from =ln to =. y y=e - -ln Figure 6..

27 INTEGAL CALCULUS - EXECISES 66 Hence, A = ln e = e ln = e e ln = =.. is the region in the first quadrant that lies below the curve y = +4 and is bounded by this curve, the line y = +,andthecoordinate ais. Solution. First sketch the region as shown in Figure 6.. Note that the curve y = +4and the line y = + intersect in the first quadrant at the point (, 8), since =is the only positive solution of the equation +4= +, i.e. + 6=. Also note that the line y = +intersects the ais at the point (, ). y y= y= Figure 6.. Observe that to the left of =, is bounded above by the curve y = +4, while to the right of =, it is bounded by the line y = +. This suggests that you break into two subregions, and, as shown in Figure 6., and apply the integral formula for area to each subregion separately. In particular, and A = Therefore, A = ( +4) = µ +4 µ ( +) = + A = A + A = = 8 +8= = 5++ =. += 8.

28 INTEGAL CALCULUS - EXECISES is the region bounded by the curves y = +5and y =,the line =,andthey ais. Solution. Sketch the region as shown in Figure 6.4. y 5 5 y= y=- Figure 6.4. Notice that the region in question is bounded above by the curve y = +5 and below by the curve y = and etends from =to =. Hence, A = [( +5) ( )] = ( +5) = µ +5 = 8+5 =. 5. is the region bounded by the curves y = and y = +4. Solution. First make a sketch of the region as shown in Figure 6.5 and find the points of intersection of the two curves by solving the equation = +4 i.e. 4= to get = and =. The corresponding points (, ) and (, ) are the points of intersection. y=- +4 y 4 y= Figure 6.5.

29 INTEGAL CALCULUS - EXECISES 68 Notice that for, thegraphofy = +4lies above that of y =. Hence, A = = [( +4) ( )] = µ + +4 ( + +4) = = =9. 6. is the region bounded by the curves y = and y =. Solution. Sketch the region as shown in Figure 6.6. Find the points of intersection by solving the equations of the two curves simultaneously to get = = ( ) = = and =. The corresponding points (, ) and (, ) are the points of intersection. y y= y= - - Figure 6.6. Notice that for, thegraphofy = lies above that of y =. Hence, A = ( ) = µ = =. (a) is the region to the right of the y ais that is bounded above by the curve y =4 and below the line y =. (b) is the region to the right of the y ais that lies below the line y =and is bounded by the curve y =4, the line y =,and thecoordinateaes.

30 INTEGAL CALCULUS - EXECISES 69 Solution. Note that the curve y =4 and the line y = intersect to the right of the y ais at the point (, ), since = is the positive solution of the equation 4 =, i.e. =. (a) SketchtheregionasshowninFigure6.7. y 4 y=4- y= Figure 6.7. Notice that for, thegraphofy =4 lies above that of y =. Hence, µ A = (4 ) = ( ) = = =. (b) Sketch the region as shown in Figure 6.8. y 4 y=4- y= Figure 6.8. Observe that to the left of =, is bounded above by the curve y =, while to the right of =, it is bounded by the line y =4. This suggests that you break into two subregions, and, as shown in Figure 6.8, and apply the integral formula

31 INTEGAL CALCULUS - EXECISES 7 for area to each subregion separately. In particular, and so A = A = = = µ (4 ) = 4 A = A + A =+ 5 = 4. = = 5, 7. is the region bounded by the curve y = and the lines y = and y = 8. Solution. First make a sketch of the region as shown in Figure 6.9 and find the points of intersection of the curve and the lines by solving the equations to get = and = 8 i.e. = and =8 = and =. y y= y= y= Figure 6.9. Then break into two subregions, that etends from =to =and that etends from =to =,asshowninfigure 6.9. Hence, the area of the region is ³ A = 7 = 8 8 = 7 6 = 7 6

32 INTEGAL CALCULUS - EXECISES 7 and the area of the region is µ A = µ = 8 6 Thus, the area of the region is the sum A = A + A = 6 = 4. = = is the region bounded by the curves y = +5 and y = Solution. First make a rough sketch of the two curves as shown in Figure 6.. You find the points of intersection by solving the equations of the two curves simultaneously +5 = = ( ) 4( ) = ( )( )( +)= to get =, = and =. y y= - +5 y= Figure 6.. The region whose area you wish to compute lies between = and =, but since the two curves cross at =, neither curve is always above the other between = and =. However, since the curve y = +5is above y = +4 7 between = and =, and since y = +4 7 is above y = +5between = and =, it follows that the area of the region between = and =,is

33 INTEGAL CALCULUS - EXECISES 7 A = = = 4 + = µ = = = = and the area of the region between =and =,is A = = = + +4 = µ = = = = 4 = 8 4 += 4. Thus, the total area is the sum Homework A = A + A =+ 4 = 4. In problems through find the area of the region.. is the triangle bounded by the line y =4 and the coordinate aes.. is the rectangle with vertices (, ), (, ), (, 5) and (, 5).. is the trapezoid bounded by the lines y = +6and =and the coordinate aes. 4. is the region bounded by the curve y =, the line =4,andthe ais.

34 INTEGAL CALCULUS - EXECISES 7 5. is the region bounded by the curve y =4, the line =,andthe ais. 6. is the region bounded by the curve y = and the ais. 7. is the region bounded by the curve y = 6 5 and the ais. 8. is the region in the first quadrant bounded by the curve y =4 and the lines y = and y =. 9. is the region bounded by the curve y = and the lines y = and y =.. is the region in the first quadrant that lies under the curve y = 6 and that is bounded by this curve and the lines y =, y =,and =8.. is the region bounded by the curve y = and the ais. (Hint: eflect the region across the ais and integrate the corresponding function.). is the region bounded by the curves y = + and y = between = and =.. is the region bounded by the curve y = e and the lines y =and =. 4. is the region bounded by the curve y = and the line y =. 5. is the region bounded by the curve y = and the line y =4. 6. is the region bounded by the curves y = 6 and y =. 7. is the region bounded by the line y = and the curve y =. 8. istheregioninthefirst quadrant bounded by the curve y = + and the lines y = 8 and y =. 9. is the region bounded by the curves y = + and y = is the region bounded by the curves y = and y = +. esults ( + ln 4) e

AP Calculus AB 2004 Scoring Guidelines AP Calculus AB 4 Scoring Guidelines The materials included in these files are intended for noncommercial use by AP teachers for course and eam preparation; permission for any other use must be sought from

PRACTICE FINAL. Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 10cm. PRACTICE FINAL Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 1cm. Solution. Let x be the distance between the center of the circle

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Partial Fractions. and Logistic Growth. Section 6.2. Partial Fractions SECTION 6. Partial Fractions and Logistic Growth 9 Section 6. Partial Fractions and Logistic Growth Use partial fractions to find indefinite integrals. Use logistic growth functions to model real-life

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1. (from Stewart, page 586) Solve the initial value problem. . (from Stewart, page 586) Solve the initial value problem.. (from Stewart, page 586) (a) Solve y = y. du dt = t + sec t u (b) Solve y = y, y(0) = 0., u(0) = 5. (c) Solve y = y, y(0) = if possible. 3. a p p e n d i g DISTANCE, CIRCLES, AND QUADRATIC EQUATIONS DISTANCE BETWEEN TWO POINTS IN THE PLANE Suppose that we are interested in finding the distance d between two points P (, ) and P (, ) in the

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Section 1-4 Functions: Graphs and Properties 44 1 FUNCTIONS AND GRAPHS I(r). 2.7r where r represents R & D ependitures. (A) Complete the following table. Round values of I(r) to one decimal place. r (R & D) Net income I(r).66 1.2.7 1..8 1.8.99 2.1

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y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx. 118 2 LINEAR AND QUADRATIC FUNCTIONS 71. Celsius/Fahrenheit. A formula for converting Celsius degrees to Fahrenheit degrees is given by the linear function 9 F 32 C Determine to the nearest degree the

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Solving Quadratic Equations by Graphing. Consider an equation of the form. y ax 2 bx c a 0. In an equation of the form SECTION 11.3 Solving Quadratic Equations b Graphing 11.3 OBJECTIVES 1. Find an ais of smmetr 2. Find a verte 3. Graph a parabola 4. Solve quadratic equations b graphing 5. Solve an application involving

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INVESTIGATIONS AND FUNCTIONS 1.1.1 1.1.4. Example 1 Chapter 1 INVESTIGATIONS AND FUNCTIONS 1.1.1 1.1.4 This opening section introduces the students to man of the big ideas of Algebra 2, as well as different was of thinking and various problem solving strategies. PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient hsn.uk.net Higher Mathematics UNIT OUTCOME 1 Polnomials and Quadratics Contents Polnomials and Quadratics 64 1 Quadratics 64 The Discriminant 66 3 Completing the Square 67 4 Sketching Parabolas 70 5 Determining

REVIEW OF ANALYTIC GEOMETRY REVIEW OF ANALYTIC GEOMETRY The points in a plane can be identified with ordered pairs of real numbers. We start b drawing two perpendicular coordinate lines that intersect at the origin O on each line. 9.3 Solving Quadratic Equations by Using the Quadratic Formula 9.3 OBJECTIVES 1. Solve a quadratic equation by using the quadratic formula 2. Determine the nature of the solutions of a quadratic equation

Ax 2 Cy 2 Dx Ey F 0. Here we show that the general second-degree equation. Ax 2 Bxy Cy 2 Dx Ey F 0. y X sin Y cos P(X, Y) X Rotation of Aes ROTATION OF AES Rotation of Aes For a discussion of conic sections, see Calculus, Fourth Edition, Section 11.6 Calculus, Earl Transcendentals, Fourth Edition, Section 1.6 In precalculus

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Worksheet 1. What You Need to Know About Motion Along the x-axis (Part 1) Worksheet 1. What You Need to Know About Motion Along the x-axis (Part 1) In discussing motion, there are three closely related concepts that you need to keep straight. These are: If x(t) represents the

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POLYNOMIAL FUNCTIONS POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a

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y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx y 1 u 2 du u 1 3u 3 C Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx.

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dy dx and so we can rewrite the equation as If we now integrate both sides of this equation, we get xy x 2 C Integrating both sides, we would have LINEAR DIFFERENTIAL EQUATIONS A first-der linear differential equation is one that can be put into the fm 1 d Py Q where P and Q are continuous functions on a given interval. This type of equation occurs

Solutions to Homework 10 Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

SOLVING EQUATIONS WITH RADICALS AND EXPONENTS 9.5. section ( 3 5 3 2 )( 3 25 3 10 3 4 ). The Odd-Root Property 498 (9 3) Chapter 9 Radicals and Rational Exponents Replace the question mark by an expression that makes the equation correct. Equations involving variables are to be identities. 75. 6 76. 3?? 1 77. 1

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1) (-3) + (-6) = 2) (2) + (-5) = 3) (-7) + (-1) = 4) (-3) - (-6) = 5) (+2) - (+5) = 6) (-7) - (-4) = 7) (5)(-4) = 8) (-3)(-6) = 9) (-1)(2) = Extra Practice for Lesson Add or subtract. ) (-3) + (-6) = 2) (2) + (-5) = 3) (-7) + (-) = 4) (-3) - (-6) = 5) (+2) - (+5) = 6) (-7) - (-4) = Multiply. 7) (5)(-4) = 8) (-3)(-6) = 9) (-)(2) = Division is

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dy dx and so we can rewrite the equation as If we now integrate both sides of this equation, we get xy x 2 C Integrating both sides, we would have Linear Differential Equations A first-der linear differential equation is one that can be put into the fm 1 d P y Q where P and Q are continuous functions on a given interval. This type of equation occurs

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15.1. Exact Differential Equations. Exact First-Order Equations. Exact Differential Equations Integrating Factors SECTION 5. Eact First-Order Equations 09 SECTION 5. Eact First-Order Equations Eact Differential Equations Integrating Factors Eact Differential Equations In Section 5.6, ou studied applications of differential

The numerical values that you find are called the solutions of the equation. Appendi F: Solving Equations The goal of solving equations When you are trying to solve an equation like: = 4, you are trying to determine all of the numerical values of that you could plug into that equation. To find a maximum or minimum: Find an expression for the quantity you are trying to maximise/minimise (y say) in terms of one other variable (x). dy Find an expression for and put it equal to 0. Solve

Solutions to Exercises, Section 5.1 Instructor s Solutions Manual, Section 5.1 Exercise 1 Solutions to Exercises, Section 5.1 1. Find all numbers t such that ( 1 3,t) is a point on the unit circle. For ( 1 3,t)to be a point on the unit circle

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LESSON EIII.E EXPONENTS AND LOGARITHMS LESSON EIII.E EXPONENTS AND LOGARITHMS LESSON EIII.E EXPONENTS AND LOGARITHMS OVERVIEW Here s what ou ll learn in this lesson: Eponential Functions a. Graphing eponential functions b. Applications of eponential

Chapter 4: Exponential and Logarithmic Functions Chapter 4: Eponential and Logarithmic Functions Section 4.1 Eponential Functions... 15 Section 4. Graphs of Eponential Functions... 3 Section 4.3 Logarithmic Functions... 4 Section 4.4 Logarithmic Properties...

Exponential Functions: Differentiation and Integration. The Natural Exponential Function 46_54.q //4 :59 PM Page 5 5 CHAPTER 5 Logarithmic, Eponential, an Other Transcenental Functions Section 5.4 f () = e f() = ln The inverse function of the natural logarithmic function is the natural eponential

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SLOPE OF A LINE 3.2. section. helpful. hint. Slope Using Coordinates to Find 6% GRADE 6 100 SLOW VEHICLES KEEP RIGHT . Slope of a Line (-) 67. 600 68. 00. SLOPE OF A LINE In this section In Section. we saw some equations whose graphs were straight lines. In this section we look at graphs of straight lines in more detail

correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were: Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that

Derivatives as Rates of Change Derivatives as Rates of Change One-Dimensional Motion An object moving in a straight line For an object moving in more complicated ways, consider the motion of the object in just one of the three dimensions

Partial Derivatives. @x f (x; y) = @ x f (x; y) @x x2 y + @ @x y2 and then we evaluate the derivative as if y is a constant. Partial Derivatives Partial Derivatives Just as derivatives can be used to eplore the properties of functions of 1 variable, so also derivatives can be used to eplore functions of 2 variables. In this

Objectives. Materials Activity 4 Objectives Understand what a slope field represents in terms of Create a slope field for a given differential equation Materials TI-84 Plus / TI-83 Plus Graph paper Introduction One of the ways

AP CALCULUS AB 2008 SCORING GUIDELINES AP CALCULUS AB 2008 SCORING GUIDELINES Question 1 Let R be the region bounded by the graphs of y = sin( π x) and y = x 4 x, as shown in the figure above. (a) Find the area of R. (b) The horizontal line CHALLENGE PROBLEMS: CHALLENGE PROBLEMS 1 CHAPTER A Click here for answers S Click here for solutions A 1 Find points P and Q on the parabola 1 so that the triangle ABC formed b the -ais and the tangent

6.3 PARTIAL FRACTIONS AND LOGISTIC GROWTH 6 CHAPTER 6 Techniques of Integration 6. PARTIAL FRACTIONS AND LOGISTIC GROWTH Use partial fractions to find indefinite integrals. Use logistic growth functions to model real-life situations. Partial Fractions Name: Systems of Equations Involving Circles and Lines Date: In this lesson, we will be solving two new types of Systems of Equations. Systems of Equations Involving a Circle and a Line Solving a system Work as the Area Under a Graph of vs. Displacement Situation A. Consider a situation where an object of mass, m, is lifted at constant velocity in a uniform gravitational field, g. The applied force is