Math 8, Exam, Study Guide Problem Solution. Compute the definite integral: 5 ( ) 7 x +3 dx Solution: UsingtheFundamentalTheoremofCalculusPartI,thevalueof the integral is: 5 ( ) 7 [ ] 5 x +3 dx = 7 ln x +3x =[7ln 5 +3(5)] [7 ln +3()] =7ln5+5 0 3 = 7ln5+
Math 8, Exam, Study Guide Problem Solution. Consider the function f(x) =x x on the interval [0, ]. Compute the trapezoid and midpoint approximations T and M. Solution: Thelengthofeachsubintervalof[0, ] is: The trapezoid approximation T is: x = b a N = 0 = T = x [f(0) + f() + f()] = [ ( 0 0 )+( )+( ) ] = [0 + + 0] = The midpoint approximation M is: [ ( ) ( )] 3 M = x f + f [( = ( ) ) ( + 3 ( ) )] 3 = 3 4 + 3 4 = 3
Math 8, Exam, Study Guide Problem 3 Solution 3. The region enclosed by the graphs of the functions y = x and y = x from x =0tox = is rotated about the y-axis. Compute the volume of the resulting solid. Solution: We will use the Shell Method to compute the volume.the formula is: V =π b a x (top bottom) dx where the top curve is y = x,thebottomcurveisy = x, theintervalis0 x. Therefore, the volume is: V =π =π 0 0 x ( x x ) dx ( x 3/ x ) dx [ =π 5 x5/ 3 x3 [ =π 5 ] 3 = π 5 ] 0
Math 8, Exam, Study Guide Problem 4 Solution 4. Compute the following integrals: sin x cos 3 xdx 4 x dx Solution: Thefirstintegraliscomputedbyrewritingtheintegralusing the Pythagorean Identity cos x +sin x =. sin x cos 3 xdx= sin x cos x cos xdx = sin x ( sin x)cosxdx = (sin x sin 4 x)cosxdx Now let u =sinx. Thendu =cosxdx and we get: sin x cos 3 xdx= (sin x sin 4 x)cosxdx = (u u 4 ) du = 3 u3 5 u5 + C = 3 sin3 x 5 sin5 x + C The second integral is computed using the u-substitution method. Let u = x. Then du = dx du = dx and x =u. Substitutingtheseintotheintegralandevaluating we get: dx = 4 x = = = ( du) 4 (u) 4 4u du 4 u du u du =arcsinu + C ( ) = arcsin x + C
Math 8, Exam, Study Guide Problem 5 Solution 5. Compute the following integrals: x dx x arctan xdx Solution: The first integral is computed using the u-substitution method. Let u = x. Then du = dx and x = u +. Substitutingtheseintotheintegralandevaluatingweget: x u + dx = du x u (u = / +u /) du = 3 u3/ +4u / + C = 3 (x )3/ +4(x ) / + C The second integral is computed using Integration by Parts. Let u =arctanx and v =. Then u = and v = x. UsingtheIntegrationbyPartsformula: x + uv dx = uv u vdx we get: arctan xdx= x arctan x x x + dx The integral on the right hand side is computed using the u-substitution u = x +. Then du =xdx du = xdx and we get: x arctan xdx= x arctan x x + dx = x arctan x x + xdx = x arctan x u du = x arctan x u du = x arctan x ln u + C = x arctan x ln(x +)+C
Math 8, Exam, Study Guide Problem 6 Solution 6. Compute the following integrals: x 3 sin ( x ) dx, x + x 6 dx Solution: The first integral is computed using the u-substitution method. Let u = x. Then du =xdx du = xdx and we get: x 3 sin ( x ) dx = x sin ( x ) xdx ( ) = u sin u du = u sin udu We now use Integration by Parts to evaluate the above integral. Let w = u and v =sinu. Then w =andv = cos u. UsingtheIntegrationbyPartsformula: wv du = wv w vdu we get: u sin udu= u( cos u) ( cos u) du = u cos u + cos udu = u cos u +sinu + C Therefore, x 3 sin ( x ) dx = u sin udu = ( u cos u +sinu)+c = u cos u + sin u + C = x cos ( x ) + sin ( x ) + C
The second integral is computed using Partial Fraction Decomposition. Factoring the denominator and decomposing we get: x + x 6 = (x +3)(x ) = A x +3 + B x Multiplying the equation by (x +3)(x ) we get: =A(x ) + B(x +3) Next we plug in two different values of x to get a system of two equations in two unknowns (A, B). Letting x = 3 andx =weget: x = 3 : =A( 3 ) + B( 3+3) A = 5 x =: =A( ) + B( + 3) B = 5 Plugging these values of A and B back into the decomposed equation and integrating we get: ( ) x + x 6 dx = 5 x +3 + 5 dx x = 5 ln x +3 + ln x + C 5
Math 8, Exam, Study Guide Problem 7 Solution 7. Compute the following integrals: x +x +3 dx, x 6 ln xdx Solution: Tocomputethefirstintegral,wewillfirstcompletethesquare in the denominator. x +x +3 dx = (x +) + dx Now let u = x +. Thendu = dx and we get: (x +) + dx = u + du Now let u = v. Thendu = dv and we get: u + du = ( ) ( dv v) + = v + dv = v + dv = arctan v + C ( ) u = arctan + C Therefore, the original integral is: x +x +3 dx = ( ) x + arctan + C The second integral is computed using Integration by Parts. Let u =lnx and v = x 6.Then u = x and v = 7 x7.usingtheintegrationbypartsformula: uv dx = uv u vdx
we get: ( ) x 6 ln xdx=(lnx) 7 x7 = 7 x7 ln x 7 x 7 x7 ) dx x 6 dx = 7 x7 ln x 49 x7 + C
Math 8, Exam, Study Guide Problem 8 Solution 8. Compute the following integrals: cos ( x ) dx, x e x dx Solution: Tobeginthesolutionofthefirstintegral,wefirstusetheu-substitution method. Let u = x.thendu = dx udu= dx and we get: x cos ( x ) dx = cos u (udu) = u cos udu We now use Integration by Parts to evaluate the above integral. Let w = u and v =cosu. Then w =andv =sinu. UsingtheIntegrationbyPartsformula: wv du = wv w vdu we get: u cos udu= u sin u sin udu u cos udu= u sin u +cosu + C Therefore, cos ( x ) dx = u cos udu =(u sin u +cosu)+c =u sin u + cos u + C = x sin ( x ) +cos ( x ) + C The second integral is computed using Integration by Parts. Let u = x and v = e x.then u =x and v = ex.usingtheintegrationbypartsformula: uv dx = uv u vdx
we get: x e x dx = x e x = x e x ( ) x ex dx xe x dx AsecondIntegrationbyPartsmustbeperformed. Letu = x and v = e x.thenu =and v = ex.usingtheintegrationbypartsformulaagainweget: x e x dx = x e x [ xex ] e x dx = x e x xex + 4 ex + C
Math 8, Exam, Study Guide Problem 9 Solution 9. Compute the area enclosed between the graphs y = x and y =3 3x. Solution: 0 - y x - -3 The formula we will use to compute the area of the region is: Area = b a (top bottom) dx where the limits of integration are the x-coordinates of the points of intersection of the two curves. These are found by setting the y s equal to each other and solving for x. y = y 3 3x = x x 3x +=0 (x )(x ) = 0 x =,x= From the graph we see that the top curve is y = x and the bottom curve is y =3 3x.
Therefore, the area is: Area = = [( x ) (3 3x) ] dx ( +3x x ) dx [ = x + 3 x ] 3 x3 = [ () + 3 () 3 ] ()3 [ () + 3 () 3 ] ()3 [ = 4+6 8 ] [ + 3 3 ] 3 = 6
Math 8, Exam, Study Guide Problem 0 Solution 0. A round hole of radius b is drilled through the center of a hemisphere of radius a (a >b). Find the volume of the portion of the sphere that remains. Solution: y 0 b a x We use the Washer Method to compute the volume. The formula is: V = π d c [ (right) (left) ] dy where the right curve is x = a y and the left curve is x = b. Thelimitsofintegration are the y-coordinates of the points of intersection of the right and left curves. We find these by setting the x s equal to each other and solving for y. x = x b = a y b = a y y = a b y = ± a b
The volume is then: a b V = π =π a b a b 0 [ ( ) a y b ] dy ( a b y ) dy =π [(a b )y 3 ] a b y3 0 [ =π (a b ) a b ( ) ] 3 a b 3 = 4π 3 (a b ) 3/