Math 113 HW #11 Solutions


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1 Math 3 HW # Solutios (a) Estimate the area uder the graph of f(x) = x from x = to x = 4 usig four approximatig rectagles ad right edpoits. Sketch the graph ad the rectagles. Is your estimate a uderestimate or a overestimate? Aswer: Sice [, 4] has legth 4, each of the four rectagles will have width 4/4 =, so the right edpoits are,, 3 ad 4. Thus, the heights of the four rectagles are f() = = f() =.44 f(3) = 3.73 f(4) = 4 = Sice each rectagle has width, the area of the first rectagle is =, the area of the secod is =, etc. Thus, we ca estimate the area uder the curve as Sice f(x) is a icreasig fuctio, this is a overestimate of the actual area. (b) Repeat part (a) usig left edpoits.
2 Aswer: The edpoits of the four subitervals are the same, though ow we re iterested i the left edpoits, which are,,, ad 3. Thus, the heights of the four rectagles are f() = = f() = = f() =.44 f(3) = Thus, the area cotaied i these rectagles is , which is a uderestimate of the actual area. 8. Use Defiitio to fid a expressio for the area uder the graph of f(x) = l x x, 3 x as a it. Do ot evaluate the it. Aswer: Sice [3, ] has legth 3 = 7, if we break this iterval up ito subitervals of equal width, each will have width x = 7/. The the area uder the graph will be give by i= f(x i ) x i= l x i x i 7
3 for ay choice of sample poits x i, where x i is i the ith subiterval. Choosig, say, the right edpoit of each as the sample poit, we ca see that x i = 3 + i 7, so the above it becomes l ( 3 + i 7 ) i 7. i= Express the it as a defiite itegral o [, ]. i= cos x i x x i Aswer: This is simply the defiitio of the defiite itegral cos x x dx.. Use the form of the defiitio of the itegral give i Theorem 4 to evaluate the itegral 4 (x + x 5) dx. Aswer: Breakig the iterval [, 4] ito subitervals of equal width, each will be of width x = 4 = 3. Moreover, the right edpoit of the ith subiterval will be x i = + i 3. Therefore, the height of the ith rectagle will be (sice we re usig right edpoits), f(x i ) = x i + x i 5 ( = + i ) 3 ( + + i 3 ) 5 = + i + i i 5 = i 9 + i. 3
4 Therefore, 4 (x + x 5) dx f(x i ) x i= i= ( i 9 ) 3 + i i= [ i i 3 ] [ i i= i= [ 7 3 i + 3 i= i 3 i= i i= ] ] i= Therefore, sice = i= i = i= i = i= ( + ) ( + )( + ), we see that the above it is equal to [ 7 ( + )( + ) ( + ) ] [ = = Therefore, 4 =. (x + x 5) dx =. ] The graph of g cosists of two straight lies ad a semicircle. Use it to evaluate each itegral (a) g(x)dx Aswer: Sice o [, ] the graph of g(x) is just a straight lie of slope comig dow from y = 4 to y =, the area is just the area of the triagle 4 = 4. Sice this area is above the xaxis, defiite itegral equals the area, so g(x)dx = 4. 4
5 (b) g(x)dx Aswer: O [, ] the graph of g(x) is a semicircle of radius lyig below the xaxis. Its area is () =. Sice it lies below the axis, the itegral is egative, so (c) g(x)dx Aswer: Sice g(x)dx = g(x)dx + g(x)dx =. g(x)dx + g(x)dx = 4 + g(x)dx, we just eed to determie g(x)dx. Sice this is a straight lie of slope goig up from the xaxis (at x = ) to y = (at x = 7), it describes a triagle of area =. Sice this area lies above the axis, g(x)dx = /, so g(x)dx = Use the result of Example 3 to compute g(x)dx = 4 + = (e x )dx. Aswer: Example 3 says that ex dx = e 3 e, we eed to use the properties of the defiite itegral to express the give itegral i terms of ex dx. Now, by Property 4, I tur, by Property, By Property 3, Puttig these together, the, (e x )dx = e x dx = (3 ) =. e x dx = (e x )dx = e x dx. Pluggig i the value we kow for ex dx, we see that dx. e x dx. (e x )dx = (e 3 e) = (e 3 e )
6 Use Part of the Fudametal Theorem of Calculus to fid the derivative of the fuctio h(x) = x + r 3 dr. Aswer: Make the chage of variables u = x. The ( ) h (x) = d x + r dx 3 dr = d dx By the Chai Rule, this is equal to d du ( u ( u + r 3 dr) du dx. ) + r 3 dr. Usig the Fudametal Theorem ad the fact that du dx = x, we see that. Evaluate the itegral h (x) = + u 3 (x) = + (x ) 3 (x) = x + x. cos θ dθ. Aswer: Sice si θ is a atiderivative of cos θ, the secod part of the Fudametal Theorem says that [ ] cos θ dθ = si θ = si si = =. 3. Evaluate the itegral Aswer: Sice we see that is a atiderivative of x. Therefore, [ x x dx = l x dx. d dx (x ) = x l, ] x l = l l = 9 l.
7 4. Evaluate the itegral Aswer: Rewrite the itegral as ( ) 4 u 3 + u u 3 du = 4 + u du. u 3 4u 3 du + u du. The, sice u = is a atiderivative for u 3 ad sice l u is a atiderivative for u u, we see that the above is equal to [ 4 ] [ ] ( u + l u = 4 + ) + (l l ) = 3 + l. 7
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