Calculus 1: Sample Questions, Final Exam

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1 Calculus : Sample Questions, Final Eam. Evaluate the following integrals. Show your work and simplify your answers if asked. (a) Evaluate integer. Solution: e 3 e (b) Evaluate integer. Solution: π π (c) Compute integer. e 3 e d. Your answer should be in the form of an e 3 d = ln = ln e 3 ln e = ln(e 3 ) ln(e ) = 3 =. e π π cos θ dθ. Your answer should be in the form of an cos θ dθ = sin θ π π = sin( π) sin( π ) = ( ) =. 3 d. Your answer should be in the form of an e Solution: d = since it is a definite integral over an 5 e + 9 interval of zero length. (This integral is probably impossible to do otherwise.) (d) Evaluate sin( 3 + ) d. Solution: Perform a substitution u = 3 +, du = 3 d, d =

2 du, and so 3 sin( 3 + ) d = sin( 3 + ) d = sin u 3 du = sin u du 3 = ( cos u) + C 3 = 3 cos(3 + ) + C. y 5 (e) Evaluate dy. Your answer should be in the form of an y + integer. Solution:. This is because it is an integral of an odd function over a symmetric interval [, ]. To see the integrand is an odd ( y) 5 function, compute ( y) + = y5 y + = y5 y +. You can also do this the long way. Use long division of polynomials to show y5 y + = y3 y + y 5 y + dy = y and integrate y + ( y 3 y + y ) dy y + = (y 3 y y) dy + y + dy = ( 4 y4 y ) + ln(y + ) = ( ) ( ) + ln ln =. 4 4 y (This computation uses the fact that y + dy = ln(y + ), which follows by using the substitution u = y +.) (f) Evaluate an integer. 3 z dz. Your answer should be in the form of

3 Solution: Make the substitution u = z, du = dz, u() =, u() = 9 to compute (g) Evaluate 3 z dz = 9 sec d. Solution: tan + C. (h) Evaluate e d. 3 u du = 3( 3 u 3 ) 9 = (9 3 3 ) = (7 ) = 5. Solution: Substitute u =, du = d, d = du to find e d = e u ( du) = e u du = e u + C = e + C.. Use geometry to evaluate the integral d. Show your work. Solution: The graph y = is the unit semicircle (for y ) centered at the origin. The area under the graph from = to = is a quarter circle. Its area is 4 πr = 4 π = π Let f be a continuous function on the interval [, ] which satisfies f() d = 5. Given this information, compute the integral Show your work and justify your answer. f(y) dy. Solution: Use substitution = y, d = dy, y =. So when =, y = = ; and when =, y = =. So upon substituting = y, we have 5 = f() d = So we may solve to find f(y) dy = f(y) dy = 5. f(y) dy. 4. Consider the following functions. Circle the one(s) which are increasing on an open interval containing =. No eplanation necessary.

4 ln sin tan Solution: Only sin and tan are increasing on an open interval containing = : ln is not even defined at =. has derivative, which is negative for < and positive for >. So is decreasing on (, ) and increasing on (, ). sin has derivative cos, and at =, cos = >. So sin is increasing near =. has derivative, which is when = (and ( ) negative for > in the domain and positive for < in the domain). So is a local ma of. The derivative of tan is sec, which is > when =. 5. Consider the function g() = + cos for < < π. (a) Find all the critical points of g() for < < π. Show your work. Hint: there are two of them. Solution: g () = sin = when sin =. This is possible only in the first and second quadrants (this is where the sine function is positive); these correspond to < < π and π < < π). To find the specific value, note that sin π = for π in the first 6 6 quadrant. The corresponding solution in the second quadrant is π π = 5π. So the only critical points are = π and = 5π (b) Classify each of the critical points you found in part (a) as a local maimum or a local minimum (or neither). Justify your answers. Solution: Apply the second derivative test: g () = cos. g ( π) = cos π = 3 <. (You can tell the sign of the solution 6 6 just by knowing that π is in the first quadrant, and thus its cosine 6 must be positive.) This means = π is a local maimum. 6

5 6. Compute the derivative d d [ ] +. Show your work. 3 Solution: Use the quotient rule [ ] d + = (3 )() ( + )(3 ) = d 3 ( 3 ) ( 3 ) 7. Consider the function h() = e + e for < <. (a) Find the interval(s) on which h() is increasing. Show your work. Solution: Compute h () = [e + e ( )] = (e e ). So h () = if (e e ) =, e = e, e =, ln(e ) = ln =, =, =. So this splits the real line up into two intervals (, ) and (, ). Check for each interval h ( ) = (e e) < (use the fact that e > ). So h () < on the interval (, ). Similarly, plug in h () = (e e ) > and so h () > on the interval (, ). Thus h is increasing on the interval (, ). (b) Show that h() is always concave up. Solution: Compute h () = (e + e ) = h(), which is clearly always positive. Therefore h is always concave up. 8. Consider the definite integral 4 d. Sketch the graph of the region whose area is given by the integral. By breaking the interval of integration into n = 4 equal subintervals, compute the right Riemann sum and simplify your answer. Draw a picture of the area represented by the right Riemann sum, and show your work.

6 Solution: For the function f() = 4, the interval [, ] and n = 4 subintervals, each subinterval has length = =. The right 4 Riemann sum is given by 4 i= Here are the pictures: f( i ) = f( ) + f() + f(3) + f() = = 5.

7 9. Compute the following its. Justify your answers (a) 3 4. Solution: It s possible to use l Hôpital s rule, since this it is type. So compute = l = l 6 = 3.

8 Alternately, we can compute = ( + + 3) + (3 4) = = = 3. ( (b) +. ) Solution: First note ( + ) = eln[(+ )] = e ln(+ ) = e ln(+ ). Then ln( + ) is of type, and so we use l Hôpital s rule to compute ln( + ) = ln( + ) All together, we have ( + = l + = + ( ) = + =. ) = e ln(+ ) = e = e. u 9 (c) u 3 u 4u + 3. Solution: This is of type, and so we use l Hôpital s rule to compute u 3 u 9 u 4u + 3 = l Alternately, we can factor u 3 u 3 u u 4 = = 3. u 9 u 4u + 3 = (u 3)(u + 3) u 3 (u 3)(u ) = u + 3 u 3 u = = 3.

9 . Consider the function p(r) = r 3 + 6r + 9r 4. (a) Find all the critical points of p(r) for r in the interval [, ]. Show your work. Solution: The critical points in this interval are r =,,. First of all, the endpoints r =, are critical points. Compute p (r) = 3r + r + 9 = if 3(r + 3)(r + ) = if r =, 3. Only r = is in our interval, so this is the only stationary critical point. (b) For which r does p(r) attain its minimum and maimum on the interval [, ]? Show your work. Solution: Compute p( ) = ( ) 3 + 6( ) + 9( ) 4 = = 6, p( ) = ( ) 3 + 6( ) + 9( ) 4 = = 8, p() = 3 + 6( ) + 9() 4 = = 46. So r = is the minimum and r = is the maimum.. (a) Sketch the graph of a function y = F () which satisfies all the following properties: F () has domain (, ). F () has a vertical asymptote at =. F () is increasing on the interval (, ). F () is concave down on the interval (, ). Solution:

10 (b) Give a formula for a function F () which satisfies all the properties listed in part (a). Justify your answer. Solution: F () = ln works: It has the correct domain. + ln = and so there is a vertical asymptote at =. F () = > for > and so F is increasing on (, ). Moreover, F () = < for > and so F is concave down on (, ). There are other possible answers. F () = also works.. (a) Consider the graph of y =. Compute the slope of the secant line between = and =. Show your work.

11 Solution: When =, y( ) = The slope of the secant line is then =. When =, y() =. y() y( ) = =. (b) Find a number c in (, ) which satisfies the conclusion of the Mean Value Theorem for f() = on the interval [, ]. Show your work. Solution: Since f() is differentiable on [, ], the Mean Value Theorem states that there is at least one c in (, ) which satisfies f (c) = f() f( ) =. To find this c, compute f () = and so we should solve = f (c) = c. This is satisfied when c =, which is true for c = and c =. Only c = is in the interval (, ), and so this is the solution we want.