Solutions to Exam 2, Math 10560

Transcription

1 Solutions to Exam, Math 6. Which of the following expressions gives the partial fraction decomposition of the function x + x + f(x = (x (x (x +? Solution: Notice that (x is not an irreducile factor. If we write the denominator in terms of irreducile factors we get f(x = x + x + (x (x + (x + since (x = (x (x +. Thus we see that the final answer should e A x + B (x + C x + + Dx + E x +. Use the Trapezoidal rule with step size x = to approximate the integral f(x where a tale of values for the function f(x is given elow. 4 x 4 f(x Solution: Using the formula for the trapezoidal rule with x= we see that 4 f(x x (f( + f( + f( + f( + f(4 = ( = 9 = 9.. Evaluate the integral xe x. Solution: First we find the indefinite integral using integration y parts: Let u = x and dv = e x so that du = and v = e x. So we have that xe x = xe x e x = xe x e x + C Then we see that xe x = lim ( = lim ( e e ( e e xe x ( = lim xe x e x = ( e = e

2 4. Compute the integral (x +. Solution: We have to e careful at the point where the function does not exist, namely x =. So we see that (x + = (x + + (x +. We work first on the part. We will solve this using u-sustitution. If (x+ we let u = x + (so du =, then the ounds change from x = to u = and x = to u =. Making the sustitution we see that ( (x + = du = lim u du u = lim ( u = lim So the integral is divergent. ( + ( = lim + =. Compute the integral π cos (cos(x sin(x. Solution: We solve this y u-sustitution. Let u = cos(x (so du = sin(x. Then the ounds of integration change from x = π to u = and from x = to u =. Making the sustitutions we get π cos (cos(x sin(x = cos (u du = sin(u = sin( ( sin( = sin( 6. Which of the following is an expression of the area of the surface formed y rotating the curve y = sin x etween x = and x = π aout the x-axis? Solution: The formula is given y a πy where in our situation a =, = π in and pulling the π out we get: π π + ( dy, y = sin(x and so dy sin(x + cos (x = cos(x. Plugging all

3 7. Find the centroid of the region ounded y y = e x, y =, x = and x =. Solution: First we note that the area of the region A is given y A = e x = e x Now, we find the centroid y finding x and y: x = A xe x, = e e = e y = A (ex For x, we solve the integral using integration y parts with u = x and dv = e x so that du = and v = e x. Then we get that xe x = xe x e x = xe x e x +C. Using this we get x = A xe x = e (xex e x = e ((e e ( = e For y we note that (e x = e x. Then we use u-sustitution with u = x so that du = and the ounds change from x = to u = and from x = to u =. Making the sustitution we get y = A (ex = e u du = (e 4 4(e (eu ( = e = e + 4(e 4. ( Thus the centroid lies at the coordinates e, e Use Euler s method with step size. to estimate y( where y(x is the solution to the initial value prolem y = (x (y x, y( =. Solution: This will require two steps in Euler s method. For step one, we know that x = and y =. Additionally, we know that h =.. We also know that x =. and x = so we can stop at step. y = y + h(x (y x = + (.(( = y = y + h(x (y x = + (.(. (. = + (. =. 9. Compute the arc length of the curve y = x from x = to x =. Solution: We see that dy = x = x. Plugging into the formula for arc length we get that arc length = + ( x ( = + x = (x + = (4 = (8 = 4

4 . Compute the integral x + x x. Solution: First we do long division dividing x into x + x. Doing this we get that x + x x = + x + x and x + x x = x + + x ( The first integral in ( is straightforward: = x + C. The second integral is otained using integration y partial fractions. By partial fractions we otain: So we have that x + x = x + (x (x + = A x + + B x x + = A(x + B(x + Plugging in x = gives B = and plugging in x = gives A =, so we see that A = and B =. Using this decomposition gives x + x = x + + x = ln x + + ln x + C Putting it all together, ( ecomes: x + x x = x + ln x + + ln x + C. Evaluate the integral ( x 8. Solution: We do this with u-sustitution. Let u = x so that x = u and hence x = ( u. Using this, we see that = ( udu. Also, the ounds of integration go from x = to u = and from x = to u =. Making the sustitution gives: ( x 8 = ( u 9 = 9 u = ( uu 8 du = (( 9 = ( 9 (u 8 u 9 du = 4. 4

5 . Find the solution to the initial value prolem ( xy y =, y( =. Solution: We can make this into a separale equation in the following way: ( xy = y + Now, separate and integrate to find the solution: and so y + dy = x y + dy = x tan (y = ln x + C To solve for C we use the initial value y( = giving us that tan ( = ln + C which implies that C = tan ( = π 4. Solving for y we get ( π y = tan 4 ln(x. Solve the initial value prolem y = x y, y( =. + x Solution: We first rewrite it as y = x +x y +x y + y x + = x x + which allows us to rewrite as Now, it is in standard form for a first-order linear differential equation with P (x = x x+ and Q(x = x+. We find the integrating factor (noting P (x = x+ = ln x + : I(x = P (x e = e (ln x+ = x +. So the final solution is given y y(x = ( I(xQ(x = ( I(x x + = x = ( x + C x + x + ( x (x + x + Using the initial value y( = tells us that = ( + C which means C =. So finally we have that y(x = x + x +