Given \(\displaystyle \pi = 2\int_{-1}^{1} \! \sqrt{1-x^2} \, \mathrm{d}x\), use properties of integrals to compute the definite integral,

\(\displaystyle A = \int_{-2}^{2} \! (x-3) \sqrt{4-x^2} \, \mathrm{d}x\)

(Source: Tom Apostol - Calculus Volume 1)

## Comments

Sort by:

TopNewestGot it finally! :D \(\displaystyle \int_ {-2}^2 (x - 3)\sqrt {4 - x^2}\, dx\)

\(=2 \displaystyle \int_{-2}^2 (x-3) \sqrt{1-\left(\frac{x}{2}\right)^2} \, dx\)

\(=4 \displaystyle \int_{-1}^1 (2 x-3) \sqrt{1-x^2} \, dx\)

\(=8 \displaystyle \int_{-1}^1 x \sqrt{1-x^2} \, dx-12 \int_{-1}^1 \sqrt{1-x^2} \, dx\)

\(=8(0)-6 \pi\) (the first integral is zero as the integrand is odd)

\(=-6 \pi\) – Danish Mohammed · 3 years, 4 months ago

Log in to reply