What is the integral from-4 to 0 of 1/(sqrt(16-x^2)) ?

The integral from-4 to 0 of 1/(sqrt(16-x^2)) is pi/2

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integral from-4 to 0 of 1/(sqrt(16-x^2))

\int_{- 4}^{0} \frac{1}{16 - x ^{2}} d x

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\int_{- 4}^{0} \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{0} 1 d u

= \int_{- \frac{π}{2}}^{0} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{- \frac{π}{2}}^{0}

Simplify

= [u]_{- \frac{π}{2}}^{0}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

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What is the integral from-4 to 0 of 1/(sqrt(16-x^2)) ?
The integral from-4 to 0 of 1/(sqrt(16-x^2)) is pi/2