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# Need help with this calculus problem

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)

Note by Danish Mohammed
3 years, 2 months ago

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Got 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)

$$=-6 \pi$$ · 3 years, 2 months ago