When is the area 1?

Calculus Level 3

For what value of $n$ is the area under the curve of $\dfrac { 1 }{ { x }^{ 2 }+n }$ and the $x$ -axis is equal to 1? Round to the nearest hundredths place.

The answer is 9.87.

1 solution

Brian Moehring
Jul 13, 2018

$1 = \int_{-\infty}^\infty \frac{1}{x^2+n}\,dx = \frac{1}{\sqrt{n}}\tan^{-1}\left(\frac{x}{\sqrt{n}}\right)\Big|_{-\infty}^\infty = \frac{\pi}{\sqrt{n}} \implies n = \pi^2 \approx \boxed{9.87}$

When is the area 1?

The answer is 9.87.

1 solution

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