Mathematics 1052, Calculus II Exam 1, April 3rd, 2010

Transcription

1 Mathematics 5, Calculus II Exam, April 3rd,. (8 points) If an unknown function y satisfies the equation y = x 3 x + 4 with the condition that y()=, then what is y? Solution: We must integrate y against to find y up to a constant. We will use substitution u=x + 4 and du=x y= x 3 x + 4 = 3 du= u We also know that when x= we have y=. Then which implies c= 7. This means = 3( + 4) /3 u /3 du= 3 u/3 + c= 3(x + 4) /3 + c y= 3(x + 4) /3 + c=6+c 7. Compute the following integrals (a) (5 points) e e vln v dv Solution: Use substitution u = ln(v). Note that ln (v) means (lnv) NOT ln(v). Our substitution indicates du = dv, and therefore dv = vdu. We must also change v the boundary points as suggested by the substitution. So, when v = e we have u = ln(e)= and when v= e=e / we have u=ln(e / )=. Then e e / vln v dv= / v u vdu= u = u / = (b) (5 points) cos (4x) Solution: We will use one of the double angle formulas cos (θ)= (+cos(θ))

2 Mathematics 5 Exam, April 3rd, This formula tells us that cos (4x)= (+cos(8x))= x + sin(8x) 8 + c= x + sin(8x) 6 Note that the last 8 in the denominator comes from the reverse chain rule. + c (c) (5 points) ln( 3) e x 4+ex Solution: We use the substitution u=e x. Observe that e x =(e x ), and therefore, it can be replaced by u. Moreover, du=e x =u and = du u. We also change the boundary points: when x = we have u = e = and when x = ln( 3) we have u=e ln(3 = 3. Then ln( 3) e x 3 4+e x = u du 3 4+u u = + u du= ( u ) arctan 3 = arctan( 3) arctan(/)= π 6 arctan(/) 3. ( points) Evaluate the following limit by interpreting it as a Riemann sum form of an integral, and then computing that integral: ( ( ( ) π π ( nπ ) lim cos + cos + +cos n n n) ) n n Solution: By looking at the x term, which is n, and the individual choice points iπ n where i=,,...,n, we see that the interval is [,]. Then the function we need to integrate is cos(πx). Therefore the limit above represents the integral cos(πx)= π sin(πx) = Page of 7

3 Mathematics 5 Exam, April 3rd, 4. (6 points) What is F (x) if F(x)= tan(x) sin(t)dt? Solution: We use the Fundamental Theorem of Calculus together with the Chain Rule to get d d sin(t)dt = sin() tan(x) sin(tan(x)) d tan(x) which gives us sin(tan(x))sec (x) 5. (5 points) Consider the area enclosed by the line y=x+, the parabola y=4 x and the x-axis. (a) Formulate (but do not compute) the area as an integral over x. Solution: As you can see from the graph above, the area must be computed by two different integrals. For that we need the intersection of the curves y = 4 x and y=x+. We obtain that point by setting these equations equal to each other. 4 x = x+ = =x + x = (x+)(x )= We need both intersection points x = and x =. We also need the intersection points of y=4 x with the x-axis: 4 x = = x=± Then the area is (x+ )+ (4 x ) (b) Formulate (but do not compute) the area as an integral over y. Solution: In order to write the integral which computes the area over y, we must view the area as a collection of horizontal (as opposed to vertical) line segments, and compute the lengths of these line segments as a function of y. For that, we need to write our graphs where x is the dependent variable and y is the independent variable. We obtain y=x+ = x=y for the first curve, while for the other curve we get y=4 x = x = 4 y = x=± 4 y Page 3 of 7

4 Mathematics 5 Exam, April 3rd, The solution with the + sign is the right half of the parabola, and the solution with the sign is the left half of the same parabola. Notice that, we need the y-coordinate of the intersection point to write the new integral. Since x=, by using either of the curves we get y=3. Then the area is 3 ( 4 y (y ))dy (c) Evaluate one of the integrals above. Solution: We will compute both integrals here for the purpose of demonstration. First the integral over x (x+)+ (4 x )= x + x + 4x x3 3 = 37 6 Now, the integral over y: here we use a substitution u = 4 y and du = dy. Also y= is replaced by 4, and y=3 is replaced by u= as boundary points. 3 ( 4 y (y ))dy= ( 4 u ((4 u) ))( du)= (u / + u )du 4 = 3 u3/ + u 4 u = (5 points) Evaluate the indefinite integral ln( x ) Solution: We will use the method of Integration By Parts: we set f = ln( x ) and dg=. Then d f = x x and g=x and ln( x )=xln( x ) x x + =xln( x ) x x Now, we will use method of Partial Fractions. For that, we first need to factorize the denominator as x = (x )(x+) and reduce the degree of the polynomial in the numerator by Euclidean long division: x = (x )+ Page 4 of 7

5 Mathematics 5 Exam, April 3rd, Then the second part of our integral is equal to x x = + ( A (x )(x+) =x+ x + B ) x+ In order to get the unknown coefficients A and B, we must solve =A(x+)+B(x ) = (A+B)= and A B= and therefore A= and B=. Thus the remaining part of our integral is x x+ =ln x ln x+ +c which makes our final answer ln( x )=xln( x ) x ln x +ln x+ +c 7. Compute the following integrals (a) (9 points) sin 7 (θ)cos 3 (θ)dθ Solution: There are two equally valid similar solutions. Here we give just one solution. We separate one of the cosine terms as use it for du=cos(θ)dθ and therefore we assume u=sin(θ). Notice that we write sin 7 (θ)cos 3 (θ)dθ= sin 7 (θ)cos (θ)cos(θ)dθ and make a subsititution u=sin(θ) we will have an extra cos (θ). This term must be rewritten in terms of sin(θ). Bu we have the simple identity cos (θ) = sin (θ). Then our integral transforms into u 7 ( u )du= (u 7 u 9 )du= u8 8 u + c= sin8 (θ) sin (θ) + c 8 (b) ( points) x 9 x Solution: The most obvious solution is by the method of Trigonometric Substitution and we use x=3sec(θ) because our integral contains x 3. Then we first have Page 5 of 7

6 Mathematics 5 Exam, April 3rd, =3sec(θ)tan(θ)dθ and x 9= 9sec (θ) 9= 9tan (θ)=3tan(θ) Now, our integral can be written as x 9 x 3tan(θ) = 3sec(θ) 3sec(θ)tan(θ)dθ=3 tan (θ)dθ =3 (sec (θ) )dθ=3tan(θ) 3θ+c Since sec(θ)= 3 x, we see θ=arcsec(x/3). Now, we draw right triangle with an inner angle θ which satisfies cos(θ) = 3 x. This implies tan(θ) = x 9 3. Then our final answer for this question is x 9 3arcsec(x/3)+c 8. ( points) Evaluate 4x 3x+ x (x ) Solution: We will use the method of Partial Fractions. We first notice that there is a repeated factor in the denominator. Therefore, the fractional function 4x 3x+ must split x (x ) as 4x 3x+ x = A (x ) x + B x + C x In order to solve for A, B and C we must first make a common denominator. This leads us to the equation 4x 3x+=Ax(x )+B(x )+Cx =(A+C)x +(B A)x B Then we immediately conclude that B=. Also B A= 3 which means A=. And finally, A+C=4 which yields C=3. Now, we can compute our integral: 4x ( 3x+ x (x ) = x x + 3 ) =ln x + + 3ln x +arbitrary constant x x Page 6 of 7

7 Mathematics 5 Exam, April 3rd, Bonus: Show that the integrals satisfy the recursion formula I n = x n e x I n = x n e x + n I n where I = e x + c for any natural number n. Now, using this recursion formula, evaluate x 3 e x Solution: For the recursion relations, we will try to compute our integrals using the method of Integration by Parts. We first identify f = x n and dg=e x. Then d f = nx n and g= e x, and therefore I n = x n e x = x n e x + n x n e x = x n e x + ni n as we wanted to show. For n= we see that I = x e x = e x = e x + c Now, using these formulas I 3 = x 3 e x + 3I = x 3 e x + 3( x e x + I )= x 3 e x 3x e x + 6I = x 3 e x 3x e x + 6( xe x + I )= x 3 e x 3x e x 6xe x 6e x + c = e x (x 3 + 3x + 6x+6)+c Page 7 of 7