Integral Calculus - Exercises

Transcription

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