The improper integral $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$ is

A) $- \infty$
B)0
C) $\pi$
D) $\frac { \pi } { 2 }$
E) $\infty$

Question

The improper integral $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$ is

A) $- \infty$
B)0
C) $\pi$
D) $\frac { \pi } { 2 }$
E) $\infty$

Quizplus · Accepted Answer

This integral represents the arc length of a quarter circle with radius 1. The integral $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$ is equivalent to the inverse sine function evaluated from 0 to 1, which gives $\frac{\pi}{2}$.

The Improper Integral ∫0111−x2dx\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x∫01​1−x2​1​dx

The Improper Integral $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$