Integrated Calculus II Exam Solutions /6/ Question Determine the following integrals: te t dt. We integrate by parts: u = t, du = dt, dv = e t dt, v = dv = e t dt = e t, te t dt = udv = uv vdu = te t ( e t )dt = te t + e t dt = te t e t = e t (t + ). te t dt = [e t (t+)] = (e () e ()) = e =.68. π sin (x) cos (x) dx. π We have an odd power of cos(x), so we substitute t = sin(x). Then we have dt = cos(x)dx and t = sin (x) = cos (x). Then the differential becomes: sin (x) cos (x) dx = sin (x)(cos (x)) cos(x)dx = t ( t ) dt = t ( t + t )dt = (t t + t 6 )dt. When x = π, we have t = sin( π) = and when x = π, we have t = sin( π =. So the integral becomes: π π = sin (x) cos (x) dx = (t t + t 6 ) dt (t t + t 6 ) dt = [ t3 3 t + t7 7 ] = ( 3 + 7 ) = 6 (3 + ) = =.389.
Question A region R is bounded by the curves: y = x 3 3x +x+ and y = x x+. Carefully sketch the region R. See the Maple solutions for the sketch. The curves meet where both equations hold at once, so where: x 3 3x + x + = x x +, x 3 x + 3x =, x(x x + 3) =, x(x )(x 3) =. So the meeting points are at x =, y =, x =, y = and x = 3, y = 7. Find the area of the region R. In the interval [, ], the cubic is above the parabola, so the area enclosed between them is: (x 3 3x + x + ) (x x + ) dx = x 3 x + 3x dx = [ x x3 3 + 3x ] = 3 + 3 = (3 6 + 8) =. In the interval [, 3], the cubic is below the parabola, so the area enclosed between them is: 3 (x x + ) (x 3 3x + x + ) dx = 3 = [ x +x3 3 3x ]3 = 8 x 3 + x 3x dx + (7 ) 3(8) 3 So the total area of the region R is: + 8 3 = 37. = 3 ( 6+ 36) = 8 3.
Question 3 Consider the following integral: J = ln(x) dx. Use the trapezoidal rule, T 6, with six intervals, to estimate J. The integrand f(x) = ln(x). The number of intervals n = 6. The endpoints are a = and b =. We have x = b a = 6 =, so n there are six intervals of width one-half. Then we have: T 6 = x (f() + f() + (f( ) ) + f(3) + f(7 ) + f() + f(9 )) (ln() + ln() + (ln( ) ) + ln(3) + ln(7 ) + ln() + ln(9 )) = = ln ( ()( )(3 )(7 )( )(9 ) ( )( ) ) = ( ) 6 ln = 3.666. Illustrate your calculation with a sketch of the curve y = ln(x), on the interval [, ] and the six relevant trapezoids. See the Maple solutions. Explain why your estimate of J either under-estimates or over-estimates the true value of the integral. We have f (x) = and f (x) = <, so the graph of f(x) is x x concave down and therefore the trapezoids lie below the curve, so give an underestimate of the true integral. Find K the maximum of the absolute value of the second derivative of the function ln(x) on the interval [, ]. We have f (x) =, which decreases on the interval [, ], so its x maximum value occurs at x = and is. So K =. Hence use the error formula: E n = K (b a) 3 to determine the maximum n possible error in your estimate of the integral. We have the maximum possible error as E 6 = ( )3 = =.6. (6 ) 6 Also calculate the integral exactly and compare with your estimates. ln(x)dx = [x ln(x)] x( ( ) x )dx = [x ln(x) x] = ln 3 = 3.6689. Here we integrated by parts, with u = ln(x), du = dx and dv = dx, x so v = x. We have an accurate estimate, the percentage error being less than one fifth of one percent. The actual error is about.63. 3
Question Determine the following integrals: J = / x x dx. We split this integral into two parts: J = J J : J = x x dx. Here we substitute: u = x, du = xdx; u range from to 3 : J = J = x dx. 3 x dx = x du u = [ u ] 3 = 3. This is a standard integral: the indefinite integral is arcsin(x). So we get: J = [arcsin(x)] = arcsin( ) arcsin() = π. 6 So the required integral is J = J J = 3 π =.698. 6 3x + x 6x + dx. We use partial fractions: we have x 6x + = (x )(x ), so the set-up for the partial fraction decomposition reads: 3x + (x )(x ) = A x + B x. Multiplying both sides of this equation by (x )(x ) gives the relation: 3x + = A(x ) + B(x ). Putting x = gives = A, so A =. Putting x = gives 6 = B, so B =. So the integral becomes: 3x + ( x 6x + dx = x ) dx x ( )] = [ ln( x ) ln( x )] [ln = (x ) x = ln( ) ln( 6 ) = ln(6 ) = ln() ln(3) 3 ln() =.3639. 3
Question A dangerous radioactive material is to be used in a medical procedure on patient X. The half-life of the material is six hours. At 6am Monday morning a sealed capsule containing grams of the material is delivered to the facility where the procedure is to be carried out. The capsule is to be administered to patient X, when grams of the material remain. When will that be? The remaining material is flushed out of patient X, when milligram of the material remains. When will that be? Since the half life is 6 hours, the amount of material at time t hours is proportional to ( ) t 6. Measuring t from 6am on Monday, since we initially have grams, the amount A(t) in grams at time t is therefore: A(t) = ( ) t 6. The time at which A(t) = is obtained by solving the equation: = ( ) t 6, = ( ) t 6, ln( ) = ln() = t 6 ln( ) = t 6 ln(), t = 6 ln() ln() = 3.93687. So at minutes and seconds after 7pm, the capsule should be administered to the patient. Finally to find when milligram is left, we solve the equation: 3 = ( ) t t t 6, = 6, ln() = 6 ln(), t = 6 ln() = 8.7678. ln() So on Thursday, 3 minutes and 3 seconds after 7pm, the material can be flushed out.
Question 6 Solve the following differential equation, with the initial condition y() = : dy dt = ( y ). yt Plot the solution and discuss its behavior as a function of t. The equation is separable, so we first separate by multiplying both sides by dt and by y and by dividing both sides by y : ydy y = dt t. Now we need to integrate both sides. We have dt = ln(at). t For the integral ydy, we substitute u = y, so du = ydy. y Then we have ydy = du = y u ln( y )+C. (The quantity y is initially positive, so we do not need an absolute value sign here ). So integrating we get: ln( y ) = ln(at), ln( y ) = ln(at) = ln((at) ), y = (At). Putting in the intitial condition, t = and y = gives 6 = A, so A = 9. Then y = A t = 9t, so y = 9t and we finally have the required solution: y = 9 t. Going back in t, from the initial value t =, we see that y goes to zero when t = 3, and the solution is not defined for t < 3. As t 3 +, the slope dy. dt Graphically, however, we can see that the solution curve y = 9 nicely t meets the solution curve y = 9 at ( 3, ). t Going forward in t from t =, we see that y gradually increases, with limit, so approaches a horizontal asymptote of y =, from below, as t. 6