int_0^1 frac{1}{sqrt{3+2x-x^2}} dx = | vedclass

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(D) To evaluate the integral $I = \int_0^1 \frac{1}{\sqrt{3+2x-x^2}} dx$,first complete the square for the quadratic expression in the denominator: $3

Vedclass · Accepted Answer

$\frac{\pi}{6}$

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(D) To evaluate the integral $I = \int_0^1 \frac{1}{\sqrt{3+2x-x^2}} dx$,first complete the square for the quadratic expression in the denominator: $3+2x-x^2 = 4 - (x^2-2x+1) = 2^2 - (x-1)^2$. Thus,the integral becomes: $I = \int_0^1 \frac{dx}{\sqrt{2^2 - (x-1)^2}}$. Using the standard integral formula $\int \frac{dx}{\sqrt{a^2-u^2}} = \sin^{-1}(\frac{u}{a}) + C$,we get: $I = \left[ \sin^{-1}\left(\frac{x-1}{2}\right) \right]_0^1$. Evaluating at the limits: $I = \sin^{-1}\left(\frac{1-1}{2}\right) - \sin^{-1}\left(\frac{0-1}{2}\right) = \sin^{-1}(0) - \sin^{-1}\left(-\frac{1}{2}\right)$. Since $\sin^{-1}(0) = 0$ and $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$,we have: $I = 0 - (-\frac{\pi}{6}) = \frac{\pi}{6}$.

$\int_0^1 \frac{1}{\sqrt{3+2x-x^2}} dx =$

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