By part 2 of the Fundamental Theorem of Calculus, $\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$ is

A) $\frac { \pi } { 6 }$
B) $\frac { \pi } { 4 }$
C) $\frac { \pi } { 2 }$
D) $\pi$
E) $\frac { 7 \pi } { 6 }$

Question

By part 2 of the Fundamental Theorem of Calculus, $\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$ is

A) $\frac { \pi } { 6 }$
B) $\frac { \pi } { 4 }$
C) $\frac { \pi } { 2 }$
D) $\pi$
E) $\frac { 7 \pi } { 6 }$

Quizplus · Accepted Answer

By the Fundamental Theorem of Calculus, Part 2, $\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x = \sin ^ { - 1 } x | _ { 0 } ^ { \frac { 1 } { 2 } } = \sin ^ { - 1 } \frac { 1 } { 2 } = \frac { \pi } { 6 }$

By Part 2 of the Fundamental Theorem of Calculus ∫01211−x2dx\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x∫021​​1−x2​1​dx

By Part 2 of the Fundamental Theorem of Calculus $\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - x ^ { 2 } } } d x$