Evaluate the improper integral \( \displaystyle\int_1^{\infty} \dfrac{1}{x^2}\, dx \).
\( \displaystyle\lim_{b\to\infty}\int_1^b x^{-2}\,dx = \lim_{b\to\infty}\Big[-\dfrac{1}{x}\Big]_1^b = \lim_{b\to\infty}\Big(-\dfrac{1}{b} + 1\Big) = 1 \).
Methods in Calculus
12 exam-style questions on improper integrals, mean value of a function and inverse trigonometric integrals. Tap Show answer for full working.
4 Easy4 Medium4 Hard
Find the mean value of \( f(x) = x^2 \) on the interval \( [0, 2] \).
Mean value \( = \dfrac{1}{2-0}\displaystyle\int_0^2 x^2\,dx = \dfrac{1}{2}\Big[\dfrac{x^3}{3}\Big]_0^2 = \dfrac{1}{2}\cdot\dfrac{8}{3} = \dfrac{4}{3} \).
Write down \( \dfrac{d}{dx}(\sin^{-1}x) \) and hence find \( \displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\, dx \).
\( \dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}} \). Therefore \( \displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\, dx = \sin^{-1}x + C \).
Evaluate \( \displaystyle\int_0^1 \dfrac{1}{1+x^2}\, dx \).
\( \Big[\tan^{-1}x\Big]_0^1 = \tan^{-1}1 - \tan^{-1}0 = \dfrac{\pi}{4} \).
Determine whether \( \displaystyle\int_0^1 \dfrac{1}{\sqrt{x}}\, dx \) converges, and if so evaluate it.
\( \displaystyle\lim_{a\to 0^+} \int_a^1 x^{-1/2}\,dx = \lim_{a\to 0^+}\Big[2x^{1/2}\Big]_a^1 = \lim_{a\to 0^+}(2 - 2\sqrt{a}) = 2 \). Converges to 2.
Find the mean value of \( f(x) = \sin x \) on \( [0, \pi] \).
Mean \( = \dfrac{1}{\pi}\displaystyle\int_0^{\pi}\sin x\,dx = \dfrac{1}{\pi}\Big[-\cos x\Big]_0^{\pi} = \dfrac{1}{\pi}(1+1) = \dfrac{2}{\pi} \).
Evaluate \( \displaystyle\int \dfrac{1}{\sqrt{4-x^2}}\, dx \).
Write \( \dfrac{1}{\sqrt{4-x^2}} = \dfrac{1}{2\sqrt{1-(x/2)^2}} \). Let \( u = x/2 \), \( du = dx/2 \).
\( \displaystyle\int \dfrac{1}{\sqrt{4-x^2}}\,dx = \sin^{-1}\Big(\dfrac{x}{2}\Big) + C \).
Show that \( \displaystyle\int_2^{\infty} \dfrac{1}{x(\ln x)^2}\, dx \) converges and find its value.
Let \( u = \ln x \), \( du = dx/x \). When \( x=2 \), \( u=\ln 2 \); as \( x\to\infty \), \( u\to\infty \).
\( \displaystyle\int_{\ln 2}^{\infty} u^{-2}\,du = \lim_{b\to\infty}\Big[-\dfrac{1}{u}\Big]_{\ln 2}^{b} = \dfrac{1}{\ln 2} \). Converges.
Find the mean value of \( f(x) = e^{-x} \) on \( [0, a] \) and determine the limit of this mean value as \( a\to\infty \).
Mean \( = \dfrac{1}{a}\displaystyle\int_0^a e^{-x}\,dx = \dfrac{1}{a}\Big[-e^{-x}\Big]_0^a = \dfrac{1 - e^{-a}}{a} \).
As \( a\to\infty \), \( \dfrac{1-e^{-a}}{a}\to 0 \).
Evaluate \( \displaystyle\int_0^{1/2} \dfrac{1}{\sqrt{1-4x^2}}\, dx \).
Let \( u = 2x \), \( du = 2\,dx \). Limits: \( x=0\Rightarrow u=0 \); \( x=1/2\Rightarrow u=1 \).
\( \displaystyle\int_0^{1/2} \dfrac{1}{\sqrt{1-4x^2}}\,dx = \dfrac12\int_0^1 \dfrac{1}{\sqrt{1-u^2}}\,du = \dfrac12\Big[\sin^{-1}u\Big]_0^1 = \dfrac12\cdot\dfrac{\pi}{2} = \dfrac{\pi}{4} \).
Determine the values of \( p \) for which \( \displaystyle\int_1^{\infty} x^{-p}\, dx \) converges.
For \( p eq 1 \): \( \displaystyle\lim_{b\to\infty}\Big[\dfrac{x^{1-p}}{1-p}\Big]_1^b = \lim_{b\to\infty}\Big(\dfrac{b^{1-p}}{1-p} - \dfrac{1}{1-p}\Big) \).
This limit is finite iff \( 1-p < 0 \) i.e. \( p > 1 \).
For \( p=1 \): \( \ln b\to\infty \), diverges. Hence converges for \( p > 1 \).
Find \( \displaystyle\int \dfrac{1}{x\sqrt{x^2-1}}\, dx \) for \( x > 1 \).
Let \( x = \sec\theta \), \( dx = \sec\theta\tan\theta\,d\theta \), \( \sqrt{x^2-1} = \tan\theta \).
Integral becomes \( \displaystyle\int \dfrac{\sec\theta\tan\theta}{\sec\theta\cdot\tan\theta}\,d\theta = \int d\theta = \theta + C = \sec^{-1}x + C \).
(Alternatively \( \cos^{-1}(1/x) + C \).)