School of Mathematics and Statistics
Carleton University
Math. 69.104
TEST 3 SOLUTIONS

This test is out of a Total of 40.

PART I: Multiple Choice Questions
(Choose and CIRCLE only ONE answer)

1. [3 marks] Which of the following expressions gives the area of the region bounded by the curves $y = x^2,$ $y = 0$, and between the lines $x=0$ and $x=1$?

   
   (a) $$I = \int_{0}^{1}{\sqrt x\ dx}$$
   $$I = \int_{0}^{1}{x^2\ dx}$$
   (c) $$I = \int_{-4}^{4}{x^2\ dx}$$
   (d) $$I = \int_{0}^{1}{\ dx}$$
   

   
   

2. [3 marks] Evaluate the integral $$\int_{0}^{1}{xe^{x^2} \ dx}$$. (a) $e$ (b) $$\frac{e}{2}$$ $$\frac{e-1}{2}$$ (d) 0.

   
   

   
   

3. [3 marks] Solve the inequality $$x^2-3x+2 < 0$$ for $x$.

   
   (a) $$\{ x : - 2 < x < -1 \}$$
   (b) $$\{ x : 2 < x < \infty \}$$
   (c) $$\{ x : -\infty < x < 1 \}$$
   $$\{ x : 1 < x < 2 \}$$
   

   
   

4. [3 marks] Evaluate the indefinite integral: $$\int{\cos (3t+1)\ dt}.$$ (a) $$\frac{\sin (4t+1)}{4} + C$$ $$\frac{\sin (3t+1)}{3} + C$$ (c) $$\sin (3t+1) + C$$ (d) $$\sin t + C$$
5. [3 marks] Answer TRUE or FALSE: For $x > 0$, the value of $$\int {\frac{dx}{x\ln x}} = \ln (\ln x ) + C$$ where $C$ is a constant.

   
   TRUE, (b) FALSE

   
   

PART II: Show all work here.
No additional pages will be accepted

1. [7+6 marks] Evaluate the following integrals using any method: a) $$\int_{0}^{\frac{\pi}{2} }{\sin^{3} x \cos x \ dx }$$. Two methods of Solution: 1) Let $u =\sin x$, $du = \cos x dx$. When $x=0, u=0$ and when $x=\pi/2$, then $u=1$. The limits get changed from $0, \pi/2$ to $0, 1$. Thus,
   $$\int_{0}^{\frac{\pi}{2} }{\sin^{3} x \cos x \ dx } = \int_{0}^{1}{u^3\ du} = \frac{1}{4}.$$
   2) Proceed as above and find an antiderivative ... Thus,
   $$\int{\sin^{3} x \cos x \ dx } = \int{u^3\ du} = \frac{u^4}{4} = \frac{\sin^4x}{4}.$$
   Then, by definition of the definite integral, we get
   $$\int_{0}^{\frac{\pi}{2} }{\sin^{3} x \cos x \ dx }= \frac{\sin^4x}{4}\bigg \vert _{0}^{\frac{\pi}{2}} = \frac{1}{4}.$$
   b) Evaluate $$\int{x^2\ 2^{2x^3+1}\ dx}$$ using any method. Solution Let $u = 2x^3+1, \ du = 6x^2 \ dx$. Then, solving for $x^2\ dx$ we get $x^2\ dx = \frac{du}{2}.$ Thus,
   $$\int{x^2\ 2^{2x^3+1}\ dx }= \frac{1}{6} \int {2^u\ du} = \frac{1}{6} \frac{2^u}{\ln 2} + C = \frac{1}{6} \frac{2^{2x^3+1}}{\ln 2} + C.$$

2. Solve the following inequality using any method: [6 marks] (a) $(x^2-1)(x-2) < 0.$ Solution We write $p(x) = (x^2-1)(x-2) = (x-1)(x+1)(x-2)$. So, the SDT looks like:

   	$x-1$	$x+1$	$x-2$	Sign of $p(x)$
   $(-\infty, -1)$	$-$	$-$	$-$	$-$
   $(-1, 1)$	$-$	$+$	$-$	$+$
   $(1,2)$	$+$	$+$	$-$	$-$
   $(2,\infty)$	$+$	$+$	$+$	$+$

   So, the solution is the set of points $x$ such that either $x < -1$ or $1 < x < 2$. [6 marks] (b) Sketch (1 mark) the graph of the function defined by $$f(x) = x^3 - 3x$$ for $x$ in the interval $(-\infty, +\infty)$. Be sure to find all critical points (2 marks), points of inflection (1 mark), concavity intervals (1 mark) and intervals of monotonicity (1 mark). Solution $f(x) = x(x^2-3) = x(x-\sqrt 3)(x+\sqrt 3)$. The zeros of $f$ are at $x = \pm \sqrt 3$ and $x=0$. Next, $f^{\prime}(x) = 3x^2-3 = 3(x^2-1) =0$ when $x=\pm 1$. Thus, $x=\pm 1$ are the only critical points (since $f^{\prime}$ always exists in this case). Furthermore, $f{\prime\prime}(x) = 6x$ and so the graph is concave up when $x > 0$ and concave down when $x < 0$. On the other hand, $f{\prime\prime}(x) = 6x =0$ only when $x=0$ and so this is indeed a point of inflection (since there is a change in concavity around $x=0$). Finally, the graph is increasing when $f^{\prime}(x) = 3x^2-3 = 3(x^2-1) > 0$, and this only happens if $\vert x\vert > 1$ and decreasing when $\vert x\vert < 1$. The graph looks like