Homework 5 - Improper Integrals

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Please enter in all answers as exact answers (e.g. $\pi$ instead of $3.1415...$ , $\sqrt {2}$ instead of $1.41...$ ). If the integral diverges, type in $INF$ if it goes towards positive infinity, $-INF$ if it goes towards negative infinity, or $DIV$ if it diverges but doesn’t go to infinity (all letters must be upper case).

1 :

Find the area $A$ under the curve $y = \frac {1}{\sqrt {1-x^2}}$ between $x = 0$ and $x = 1$ .

$A = \answer {\frac {\pi }{2}}$

2 :

Compute $\displaystyle \int _{-2}^2 \frac {1}{x^2}dx$ .

$\displaystyle \int _{-2}^2 \frac {1}{x^2}dx = \answer {INF}$

3 :

Find a positive value of $a$ such that $\displaystyle \int _0^{\infty } e^{-ax}dx = 5$ .

$a = \answer {\frac {1}{5}}$

4 :

Calculate $\displaystyle \int _0^8 \frac {dx}{x^{1/3}}$ .

$\displaystyle \int _0^8 \frac {dx}{x^{1/3}} = \answer {6}$

5 :

Calculate $\displaystyle \int _0^{\pi /2} \sec ^2(t)dt$ .

$\displaystyle \int _0^{\pi /2} \sec ^2(t)dt = \answer {INF}$

6 :

Find $\displaystyle \int _{-\infty }^0 xe^x \:dx$ .

$\displaystyle \int _{-\infty }^0 xe^x \:dx = \answer {-1}$

7 :

Calculate $\displaystyle \int _0^{\infty } \frac {2}{x^2+2x+2}dx$ .

$\displaystyle \int _0^{\infty } \frac {2}{x^2+2x+2}dx = \answer {\frac {\pi }{2}}$

8 :

Given that $\displaystyle \int _0^{\infty } e^{-x^2}\:dx = \sqrt {\pi }/2,$ find $\displaystyle \int _0^{\infty }e^{-a^2x^2}\:dx$ , where $a$ is a negative constant (your answer will be in terms of $a$ ).

$\displaystyle \int _0^{\infty }e^{-a^2x^2}\:dx = \answer {\frac {\sqrt {\pi }}{-2a}}$

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