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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$.We know that the antiderivative of $\frac{1}{\sqrt{1-x^{2}}}$ is $\sin^{-1} x$.Let $F(x) = \sin^

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) Let $I = \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$.We know that the antiderivative of $\frac{1}{\sqrt{1-x^{2}}}$ is $\sin^{-1} x$.Let $F(x) = \sin^{-1} x$.By the second fundamental theorem of calculus,we have $I = F(1) - F(0)$.Substituting the limits,we get $I = \sin^{-1}(1) - \sin^{-1}(0)$.Since $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(0) = 0$,we have $I = \frac{\pi}{2} - 0$.Therefore,$I = \frac{\pi}{2}$.

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

Similar Questions

$\int_0^2 \frac{2x-2}{2x-x^2} dx$ is equal to

If the value of the integral $\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} \,dx = \alpha e^{-1} + \beta$,where $\alpha, \beta \in R, 5\alpha + 6\beta = 0$,and $[x]$ denotes the greatest integer less than or equal to $x$; then the value of $(\alpha + \beta)^{2}$ is equal to:

If $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$,then $\lim_{t \rightarrow \infty} f(t)$ is equal to:

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

Evaluate the definite integral $\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module