What is the integral from 2 to 4 of 1/(sqrt(16-x^2)) ?

The integral from 2 to 4 of 1/(sqrt(16-x^2)) is pi/3

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integral from 2 to 4 of 1/(sqrt(16-x^2))

\int_{2}^{4} \frac{1}{16 - x ^{2}} d x

\frac{π}{3}

Decimal

1.04719 \dots

Solution steps

\int_{2}^{4} \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{\frac{π}{6}}^{\frac{π}{2}}

Simplify

= [u]_{\frac{π}{6}}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{3}

= \frac{π}{3}

\frac{\partial}{\partial y} (e^{x^{2}})

\int 4 x^{- 5} d x

\int (x - 1) e^{- x - 1} d x

n = 0 \sum \infty e

f (x) = x - \frac{1}{x}

What is the integral from 2 to 4 of 1/(sqrt(16-x^2)) ?
The integral from 2 to 4 of 1/(sqrt(16-x^2)) is pi/3