18.01 Practice Exam 3

1. (10 points each) Find the following indefinite integrals.

(a) ∫ ( ^{3} \sqrt{x} + 1)^{2} dx. Solution:

\begin{aligned}∫ ( ^{3} \sqrt{x} + 1)^{2} dx & = ∫ (x^{2/3} + 2x^{1/3} + 1) dx \\ & = \frac{3}{5}x^{5/3} + \frac{3}{2}x^{4/3} + x + C. \end{aligned}

(b) ∫ \frac{x}{(x^{2} + 1)^{2}} dx.

Solution: Let u = x^{2} + 1, so du = 2x dx and

∫ \frac{2x}{2(x^{2} + 1)^{2}} dx = ∫ \frac{1}{2u^{2}} du

= ∫ \frac{1}{2}u^{−2} du

= \frac{−1}{2}u^{−1} + C

= \frac{−1}{2}(x^{2} + 1)^{−1} + C

= \frac{−1}{2(x^{2} + 1)} + C.

(c) ∫ (1 + cos^{3} x) sin x dx.

Solution: Let u = cos x, so du = − sin x dx and

∫ (1 + cos^{3} x) sin x dx = ∫ −(1 + u^{3}) du

= −(u + \frac{1}{4}u^{4}) + C

= − cos x − \frac{1}{4} cos^{4} x + C.

2. (11 points each) Evaluate the following definite integrals. (a) ∫_{0}^{1}(e^{x} + x^{e}) dx.

Solution:

∫_{0}^{1}(e^{x} + x^{e}) dx = (e^{x} + \frac{1}{e + 1}x^{e+1})]_{0}^{1}

= (e + \frac{1}{e + 1}) − 1

= \frac{(e − 1)(e + 1)}{e + 1} + \frac{1}{e + 1}

= \frac{e^{2}}{e + 1}.

(b) ∫_{0}^{π/4} sin(2x) dx.

Solution: Since ∫ sin(2x) dx = −\frac{1}{2} cos(2x) + C, we have

∫_{0}^{π/4} sin(2x) dx = −\frac{1}{2} cos(2x)]_{0}^{π/4}
= −\frac{1}{2} cos(π/2) + \frac{1}{2} cos 0
= \frac{1}{2}.

3. (12 points) If a > 0, then y = a − x^{4} has one arch above the x-axis. Find its area.

Solution: The graph crosses the axis when x^{4} = a, so x = ± ^{4} \sqrt{a}. The graph is above the x-axis when x is between −a^{1/4} and a^{1/4}, so the area we want is given by

∫_{−a^{1/4}}^{a^{1/4}}(a − x^{4}) dx = (ax − \frac{1}{5}x^{5})]_{−a^{1/4}}^{a^{1/4}}

= (a · a^{1/4} − \frac{1}{5}(a^{1/4})^{5}) − (a · (−a^{1/4}) − \frac{1}{5}(−a^{1/4})^{5})

= a^{5/4} − \frac{1}{5}a^{5/4} + a^{5/4} − \frac{1}{5}a^{5/4}
= \frac{8}{5}a^{5/4}.

4. (12 points) Find the lower sum for ∫_{1}^{2} \frac{1}{x} dx with n (equal) intervals.

Solution: The intervals have width \frac{1}{n}, so their endpoints, along with the corresponding values of y are:

x_{0} = 1,  y_{0} = 1
x_{1} = 1 + \frac{1}{n} = \frac{n+1}{n}, y_{1} = \frac{n}{n+1}
x_{2} = 1 + \frac{2}{n} = \frac{n+2}{n}, y_{2} = \frac{n}{n+2}
⋯
x_{n} = 2,  y_{n} = \frac{1}{2}.

The minimum value of 1/x occurs at the right endpoint of each interval, so the lower sum is:

\frac{1}{n} (\frac{n}{n + 1} + \frac{n}{n + 2} + ⋯ + \frac{1}{2}) = \frac{1}{n + 1} + \frac{1}{n + 2} + ⋯ + \frac{1}{2n}.

5. (12 points) Use Simpson’s Rule with n = 4 to approximate the value of

∫_{0}^{2} \frac{1}{4x^{2} + 1} dx.

Solution: The values of x and y at endpoints are:

\begin{matrix}x_{0} = 0, & y_{0} = 1 \\ x_{1} = \frac{1}{2}, & y_{1} = \frac{1}{2} \\ x_{2} = 1, & y_{2} = \frac{1}{5} \\ x_{3} = \frac{3}{2}, & y_{3} = \frac{1}{10} \\ x_{4} = 2, & y_{4} = \frac{1}{17}. \end{matrix} .

Simpson’s Rule gives

∫_{0}^{2} \frac{1}{4x^{2} + 1} dx ≈ \frac{1}{6} (1 + 4 · \frac{1}{2} + 2 · \frac{1}{5} + 4 · \frac{1}{10} + \frac{1}{17})
= \frac{1}{6}(3 + \frac{4}{5} + \frac{1}{17})
= \frac{1}{6} · \frac{255+68+5}{85} = \frac{328}{510} = \frac{164}{255}.

6. (12 points) Determine the intervals where the function

f(x) = ∫_{0}^{x}(e^{t^{2}} − 2) dt

is increasing.

Solution: By the (Second) Fundamental Theorem of Calculus,

f′(x) = e^{x^{2}} − 2.

This is positive when e^{x^{2}} > 2, or equivalently, when x^{2} > ln 2, or equivalently, when

x < −\sqrt{ln 2}, or x > \sqrt{ln 2}.

So f(x) is increasing on those two intervals.