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

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

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

\int_{- 4}^{0} (16 - x^{2}) d x

4 π

Decimal

12.56637 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{0} 16 cos^{2} (u) d u

= \int_{- \frac{π}{2}}^{0} 16 cos^{2} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int_{- \frac{π}{2}}^{0} cos^{2} (u) d u

Rewrite using trig identities

= 16 \cdot \int_{- \frac{π}{2}}^{0} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{0} 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{0} 1 d u + \int_{- \frac{π}{2}}^{0} cos (2 u) d u)

\int_{- \frac{π}{2}}^{0} 1 d u = \frac{π}{2}

\int_{- \frac{π}{2}}^{0} cos (2 u) d u = 0

= 16 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

16 \cdot \frac{1}{2} (\frac{π}{2} + 0) : 4 π

= 4 π

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\int (x^{2} sin (3 x)) d x

\int \frac{1}{x ^{2} 1 - \frac{1}{x ^{2}}} d x

y^{2} = 5 x^{4} - x^{2}, (1, 2)

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