t 2 + 2t dt = (t + 1) dt + 1 = arctan t x + 6 x(x 3)(x + 2) = A x +

Transcription

1 MATH 06 0 Practice Exam #. (0 points) Evaluate the following integrals: (a) (0 points). t +t+7 This is an irreducible quadratic; its denominator can thus be rephrased via completion of the square as a sum of two squares, which under appropriate division becomes an expression of the form u +, which can be integrated using an implicit linear substitution: t + t + 7 t + t (t + ) + ( t+ 6 ) + ( t+ ) + 6 arctan t + arctan t + (b) (0 points) x+6 x(x )(x+) ds. For this factorization into distinct linear terms, the appropriate decomposition is x(x )(x + ) A x + B x + + which, on multiplying by the common denominator, yields C x A(x )(x + ) + Bx(x ) x(x + ) A(x x 6) + B(x x) (x + x) 0x + (A + B )t + ( A B )t 6A Comparing quadratic, linear, and constant terms on the left and right side of the above equation yields the system of equations A + B 0 A B 6A 6 The last equation gives us A immediately; combined with the first two, we see that B and B 0, which can be reduced to B and C. Thus x(x )(x + ) x + x + + x so that x(x )(x + ) dx x + x + + x dx ln x + ln x+ + ln x +C Page of January, 0

2 MATH 06 0 Practice Exam #. (0 points) Evaluate the following integrals: (a) (0 points) t t. The term t suggests a trigonometric substitution, and in particular a sine substitution. If we construct a right triangle with marked angle θ, hypotenuse length, opposite side length t, and adjacent side length t, we find that t sin θ and t cos θ. Furthermore, for the substitution, we will need the differential cos θdθ. Then the above integral becomes ( sin θ) ( cos θ)( cos θ)dθ sin θ cos θdθ Since the above integral has an odd number of multiplicative sinθ terms, judicious use of the identity sin θ ( cos θ) will give an expression solely in terms of cos θ save for a single multiplicative term sin θ: sin θ cos θdθ sin θ( cos θ) cos θdθ (cos θ cos θ) sin θdθ This form is now ripe for the substitution u cos θ, du sin θdθ, which will serve to simplify this integral: (cos θ cos θ) sin θdθ (u u )( du) u u ( cos θ) ( cos θ) t t (b) (0 points) y y dy. +9 The term y + 9 suggests a trigonometric substitution, and in particular a tangent substitution. If we construct a right triangle with marked angle θ, hypotenuse length y + 9, adjacent side length, and opposite side length y, we find that y tan θ and t + 9 sec θ. Furthermore, for the substitution, we will need the differential sec θdθ. Then the above integral becomes ( tan θ) ( sec θ) ( sec θdθ) sec θ tan θdθ which is actually pretty easily solved: sec θdθ sec θ y + 9 Note that this problem could also be solved, and arguably much more simply, by using a standard substitution, since y + 9, in addition to being a trigonometric form, is a composition of a square root with the expression y + 9, and the derivative of y + 9 Page of January, 0

3 MATH 06 0 Practice Exam # appears in a modified form as a factor of the integrand, suggesting the substitution u y + 9, which gives du ydy, allowing us to rephrase the integral like so: y du y + 9 dy u/ u / y + 9. (0 points) The region shown below is the area between the curves y x and y x. (a) ( points) Find the area of this region. The region is bounded on the left by x and on the right by x. The upper curve is y x and the lower curve is y x. Thus, the integral to calculate the area is (x ) x dx x x x ] (8 9 9) ( ) (b) ( points) Find the volume of the solid produced by rotating this region around the x-axis. The bounds of the region are as given above; in this case, spinning the region around the x-axis, the vertical cross-sections whose height formed the integrands above trace out washers of outer radius x and inner radius x. Thus the integral to give the volume of the solid is π (x ) π ( x ) dx π 6x x + 9 x dx. (0 points) Evaluate the following integrals: ] [ 6x π x + 9x x 6 7 π ( ) 8π (a) (0 points) (x x)e x dx. This is a product of a polynomial with the integrable function e x, so we will integrate by parts, making use of the decomposition u x x, dv e x dx, which gives du (x )dx and v e x. Integration by parts thus gives: (x x)e x dx (x x)e x (e x )(x )dx The new integral we have at the end of this line is likewise a product of a polynomial and an integrable function, so again we use integration by parts, with the decomposition Page of January, 0

4 MATH 06 0 Practice Exam # u x, dv e x dx, which gives du dx and v e x. Continuing the integration by parts: (x x)e x dx (x x)e x (x )e x dx ( ) (x x)e x (x )e x e x (dx) ( ) (x x)e x (x )e x e x dx (x x)e x ((x )e x e x ) (x x + )e x (b) (0 points) x sin xdx This is a product of a polynomial with the integrable function sin x, so we will integrate by parts, making use of the decomposition u x, dv sin xdx, which gives du dx and v cos x. Integration by parts thus gives: x sin xdx x cos x x cos x + cos xdx x cos x + sin x cos x(dx). ( points) Evaluate the following integrals: (a) ( points) sin xe cos x dx. There is a composition in this integrand: e cos x naturally decomposes as a composition of the exponential function and the expression cos x; furthermore, the derivative of cos x appears in slightly modified form as a factor of the integrand, so the substitution u cos x is strongly suggested. Letting u cos x, it follows that du sin xdx. Converting the integral above using this substitution thus yields: sin xe cos x e u ( du) e u e cos x (b) ( points) tan 6 θ sec θdθ. There is a composition in this integrand: tan 6 θ naturally decomposes as a composition of the sixth power and the expression tan θ; furthermore, the derivative of tan θ appears as a factor of the integrand, so the substitution u tan θ is strongly suggested. Letting u tan θ, it follows that du sec θdθ. substitution thus yields: tan 6 θ sec θdθ Converting the integral above using this u 6 du u7 7 tan7 θ 7 Page of January, 0

5 MATH 06 0 Practice Exam # (c) ( points) x e (x ). There is a composition in this integrand: e (x) naturally decomposes as a composition of the exponential function and the expression x ; furthermore, the derivative of x appears in slightly modified form as a factor of the integrand, so the substitution u x is strongly suggested. Letting u x, it follows that du x dx, or more ot our purpose, du x dx. Converting the integral above using this substitution thus yields: x e x x x e u du ] ] x eu ex e6 e x 6. ( points) The region shown below is the area under the curve y x x from x to x. (a) ( points) Construct, but do not evaluate, an integral representing the volume of the solid produced by rotating this region around the x-axis. Working from the left bound (x ) to the right bound (x ) in the horizontal direction, the cross-sections of this solid will be discs of radius x x, so the integral setup is π(x x) dx (b) ( points) Construct, but do not evaluate, an integral representing the volume of the solid produced by rotating this figure around the y-axis. This cannot be set up as a disc problem: doing so would require inverting the function f(x) x + x, a matter of some difficulty. Thus we still need to integrate with respect to x, and thus the vertical cross-sections at particular x-values revolve out into cylindrical shells of radius x and height x x. Thus, the volume is represented by the integral πx(x x)dx (c) ( points) Calculate the average value of the function f(x) x x on the interval [, ]. The average value is x x [ ] x x/ / 6 8 / / (6 6) ( ) Page of January, 0