Exercises 2

This page allows you to practice some exercises on Integration. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 10.3 - Properties of Definite Integrals - Section 10.5 - Integration by Parts

Question 10

Evaluate the following integrals.

(a) $\int_{-2}^{2} (e^{x} - e^{-x}) \text{ d}x$

(b) $\int_{2}^{10} \frac{1}{x-1} \text{ d}x$

(c) $\int_{0}^{1} 2xe^{x^{2}} \text{ d}x$

(d) $\int_{1}^{2} \frac{1 + x^{3}}{x^{2}} \text{ d}x$

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(a) $\Big|_{-2}^{2} (e^{x} + e^{-x}) = (e^{2} + e^{-2}) - (e^{-2} + e^{2}) = 0$.

(b) $\Big|_{2}^{10} \ln(x-1) = \ln(9) - \ln(1) = \ln 9$.

(c) Let $u = x^2 \implies \frac{\text{d}u}{\text{d}x} = 2x \implies \text{d}u = 2x \text{ d}x$. This is very useful, because the integral can now be represented as

$\int_{0}^{1} e^{u} \text{ d}u = \Big|_{0}^{1} e^{u} = e^{1} - e^{0} = e - 1$,

where the bounds have been adapted to $u$ instead of $x$, but remained the same as

• $x=0 \implies u=0;$

• $x=1 \implies u = 1$.

(d) $\int_{1}^{2} \frac{1+x^3}{x^2} \text{ d}x = \int_{1}^{2} \left( \frac{1}{x^2} + x \right) \text{ d}x = \Big|_{1}^{2} \left( -\frac{1}{x} + \frac{1}{2}x^{2} \right) = 2$.

Question 11

Solve the following problems.

(a) If $\int_{a}^{b} f(x) \text{ d}x = 8$ and $\int_{a}^{c} f(x) \text{ d}x = 4$, what is $\int_{c}^{b} f(x) \text{ d}x$?

(b) If $\int_{0}^{1} (f(x) - 2g(x)) \text{ d}x = 6$ and $\int_{0}^{1} (2f(x) + 2 g(x)) \text{ d}x = 9$, find $I = \int_{0}^{1} (f(x) - g(x)) \text{ d}x$.

(c) Find the function $f(x)$ if $f'(x) = ax^{2} + bx$, $f'(1) = 6$, $f''(1) = 18$ and $\int_{0}^{2} f(x) \text{ d}x = 18$.

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(a) $\int_{c}^{b} f(x) \text{ d}x = \int_{a}^{b} f(x) \text{ d}x - \int_{a}^{c} f(x) \text{ d}x = 8 - 4 = 4$.

(b) Let $A = \int_{0}^{1} f (x) \text{ d}x$ and $B = \int_{0}^{1} g(x) \text{ d}x$. Then the two equations imply that $A − 2B = 6$ and $2A + 2B = 9$. Solving these gives $A = 5$ and $B = −1/2$, so $I = A − B = 11/2$.

(c) $f(x) = 4x^3 - 3x^2 + 5$, because:

• $f' (1) = a + b = 6$.

• $f''(1) = 2a + b = 18$.

• These are two equations with 2 unknowns that solves for $a=12$ and $b=-6$.

• Thus, $18 = \int_{0}^{2} 4x^3 - 3x^2 + C = \Big|_{0}^{2} (x^4 - x^3 + Cx) = 2^4 - 2^3 + 2C$, which solves for $C = 5$.

Question 12

Evaluate the following integrals.

(a) $\int_{0}^{3} \left( \frac{1}{3}e^{3x - 2} + (x+2)^{-1} \right) \text{ d}x$

(b) $\int_{0}^{1} \frac{x^2 + x + \sqrt{x+1}}{x+1} \text{ d}x$

(c) $\int_{1}^{b} \left( A \frac{x + b}{x + c} + \frac{d}{x} \right) \text{ d}x$ for $A, b, c, d$ positive constants

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(a) Using the integration rules and $\ln(a) - \ln(b) = \ln(a/b)$, we obtain:$\Big|_{0}^{3} \left[ \frac{1}{9} e^{3x-2} + \ln(x+2) \right] = \left[\frac{1}{9}e^{7} + \ln(5)\right] - \left[ \frac{1}{9}e^{-2} + \ln(2) \right] = \frac{1}{9}(e^7 - e^{-2}) + \ln(5/2)$.

(b) We can rewrite $\frac{x^2 + x + \sqrt{x+1}}{x+1}$ as $x + (x+1)^{-1/2}$. Note that

• The first integral is straightforward: $\int_{0}^{1} x \text{ d}x = \Big|_{0}^{1} \frac{x^2}{2} = \frac{1}{2}$.

• For the second integral, define $u = x+1$. This gives $\frac{\text{d}u}{\text{d}x} = 1$, thus $\text{d}u = \text{d}x$. When $x=0, u=1$ and when $x=1, u=2$. We now obtain $\int_{0}^{1} (x+1)^{-1/2} \text{ d}x = \int_{1}^{2} u^{-1/2} \text{ d}u = \Big|_{1}^{2} 2\sqrt{u} = 2(\sqrt{2} - 1)$.

Thus, the final result is $\frac{1}{2} + 2(\sqrt{2} - 1) = 2\sqrt{2} - \frac{3}{2}$.

(c) The final result is: $A(b-1) + A(b-c)\ln[(b+c)/(1+c)] + d\ln b$. You can get there by recognizing:

• $\int_{1}^{b} \left( A \frac{x + b}{x + c} + \frac{d}{x} \right) \text{ d}x = A\int_{1}^{b} \frac{x+b}{x+c} \text{ d}x + d \int_{1}^{b} \frac{1}{x} \text{ d}x$ (split the integral).

• Note that $\frac{x+b}{x+c} = \frac{(x+c) + (b-c)}{x+c} = 1 + \frac{b-c}{x+c}$ (rewrite the fraction).

Now, we can evaluate the following integrals:

• $\int_{1}^{b} \frac{1}{x} \text{ d}x = \ln b$, as $\ln 1 = 0$.

• $\int_{1}^{b} 1 = b-1$.

• $\int_{1}^{b} \frac{1}{x+c} = \Big|_{1}^{b} \ln(x+c) = \ln(b + c) - \ln(1 +c) = \ln[(b+c)/(c+1)]$.

Adding all the parts together (including the constants), yields the desired result.

Question 13

Let $F(x) = \int_{0}^{x} (t^2 + 2) \text{ d}t$ and $G(x) = \int_{0}^{x^{2}} (t^2 + 2) \text{ d}t.$ Find $F'(x)$ and $G'(x)$.

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Since the derivative of the definite integral with respect to the upper limit of integration is equal to the integrand evaluated at that limit, i.e.,

$\frac{\text{d}}{\text{d}t} \int_{a}^{t} f(x) \text{d} x = F'(t) = f(t)$,

we know that $F'(x) = x^2 + 2$. Combined with the chain rule, it implies that $G'(x) = G' (x^2) \times \frac{\text{d}x^2}{\text{d}x} = [(x^2)^2 + 2] \times 2x = 2x^5 + 4x$.

Question 14

Find the following expressions.

(a) $\frac{\text{d}}{\text{d}t} \int_{t}^{3} e^{-x^{2}} \text{ d}x$

(b) $\frac{\text{d}}{\text{d} t} \int_{-t}^{t} \frac{1}{\sqrt{x^{4} + 1}} \text{ d}x$

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(a) $-e^{-t^2}$

(b) $2/\sqrt{t^4 +1}$

By using the following rule, these results are immediate. If $a(t)$ and $b(t)$ are differentiable and $f(x)$ is continuous,

$\frac{\text{d}}{\text{d}t} \int_{a(t)}^{b(t)} f(x) \text{d}x = f(b(t))b'(t) - f(a(t))a'(t)$.

Question 15

Use integration by parts to evaluate the following:

(a) $\int x e^{-x} \text{ d}x$

(b) $\int 3x e^{4x} \text{ d}x$

(c) $\int (1+x^2)e^{-x} \text{ d}x$

(d) $\int x \ln x \text{ d}x$

(e) $\int_{-1}^{1} x \ln (x+2) \text{ d}x$

(f) $\int_{0}^{2} x 2^{x} \text{ d}x$

(g) $\int_{0}^{1} x^{2}e^{x} \text{ d}x$

(h) $\int_{0}^{3} x\sqrt{1 + x} \text{ d}x$

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Integration by Parts can be applied using the following formula:

$\int f(x) g'(x) \text{ d}x = f(x)g(x) - \int f' (x)g(x) \text{ d}x$.

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(a) Choose $f(x) = x$ and $g' (x) = e^{-x}$. Then we obtain

$\int xe^{−x} \text{ d}x = x(−e^{−x}) − \int 1 \cdot (−e^{−x}) \text{ d}x = −xe^{−x} − e^{−x} + C$.

(b) $\frac{3}{4}xe^{4x} - \frac{3}{16}e^{4x} + C$.

(c) $-x^2 e^{-x} - 2x e^{-x} - 3e^{-x} + C$.

(d) $\frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 + C$.

(e) $\int_{-1}^{1} x \ln (x+2) \text{ d}x = \Big|_{-1}^{1} \frac{1}{2}x^2 \ln(x+2) - \int_{-1}^{1} \frac{1}{2}x^2 \frac{1}{x+2} \text{ d}x$

$= \frac{1}{2} \ln 3 - \frac{1}{2}\int_{-1}^{1} (x-2 + \frac{4}{x+2}) \text{ d}x = 2 - \frac{3}{2}\ln 3$.

(f) $8/(\ln 2) - 3/(\ln 2)^2$.

(g) $e - 2$.

(h) $7\frac{11}{15} = \frac{116}{15}$.

Question 16

Note that we can always write $f(x) = 1 \cdot f(x)$ for any function $f(x)$. Use this fact and integration by parts to prove that $\int f(x) \text{ d}x = xf(x) - \int x f'(x) \text{ d}x$. Apply this formula to $f(x) = \ln x$.

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This result follows directly by defining $f (x) = f(x) \cdot g'(x)$, where $g'(x) = 1$. Filling in the formula $\int f(x) g'(x) \text{ d}x = f(x)g(x) - \int f' (x)g(x) \text{ d}x$ then yields the desired result $\int f(x) \text{ d}x = x \cdot f(x) - \int x f'(x) \text{ d}x$.

For $f(x) = \ln x$, we now obtain $\int \ln x \text{ d}x = x\ln x - \int x \frac{1}{x} \text{ d}x = x \ln x - x + C$.

Question 17

Evaluate the following integrals for $r \neq 0$:

(a) $\int_{0}^{T} bt e^{-rt} \text{ d}t$

(b) $\int_{0}^{T} (a + bt)e^{-rt} \text{ d}t$

(c) $\int_{0}^{T} (a - bt + ct^{2})e^{-rt} \text{ d}t$

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(a) $br^ {−2} \left[1 − (1 + rT )e^{−rT} \right]$.

(b) $ar^{−1} (1 − e^{−rT}) + br^{−2}\left[1 − (1 + rT )e^{−rT}\right]$.

(c) $ar^{−1}(1 − e^{−rT}) − br^{−2}\left[1 − (1 + rT )e^{−rT}\right]$ $\ \ \ + \ c r^{−3}\left[2 (1 − e^{−rT}) − 2rTe^{−rT} − r^2 T^2 e^{−rT}\right]$.