Evaluate the definite integral int_{0}^{1} fr | vedclass

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(A) Let $I = \int_{0}^{1} \frac{d x}{1+x^{2}}$.We know that the antiderivative of $\frac{1}{1+x^{2}}$ is $	an^{-1}(x)$.Let $F(x) = 	an^{-1}(x)$.By t

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$\frac{\pi}{4}$

Vedclass · Answer

(A) Let $I = \int_{0}^{1} \frac{d x}{1+x^{2}}$.We know that the antiderivative of $\frac{1}{1+x^{2}}$ is $	an^{-1}(x)$.Let $F(x) = 	an^{-1}(x)$.By the second fundamental theorem of calculus,we have:$I = F(1) - F(0)$$I = 	an^{-1}(1) - 	an^{-1}(0)$Since $	an^{-1}(1) = \frac{\pi}{4}$ and $	an^{-1}(0) = 0$,we get:$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

Evaluate the definite integral $\int_{0}^{1} \frac{d x}{1+x^{2}}$.

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