What is the integral from 0 to 4 of 3tsqrt(16-t^2) ?

The integral from 0 to 4 of 3tsqrt(16-t^2) is 64

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integral from 0 to 4 of 3tsqrt(16-t^2)

\int_{0}^{4} 3 t 16 - t^{2} d t

64

Solution steps

\int_{0}^{4} 3 t 16 - t^{2} d t

Take the constant out:

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

= 3 \cdot \int_{0}^{4} t 16 - t^{2} d t

Apply u-substitution

: \int_{16}^{0} - \frac{u}{2} d u

= 3 \cdot \int_{16}^{0} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 3 (- \int_{0}^{16} - \frac{u}{2} d u)

Take the constant out:

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

= 3 (- (- \frac{1}{2} \cdot \int_{0}^{16} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

= 3 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}))

Simplify

3 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16})) : \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

= \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

Compute the boundaries:

\frac{128}{3}

= \frac{3}{2} \cdot \frac{128}{3}

Simplify

= 64

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What is the integral from 0 to 4 of 3tsqrt(16-t^2) ?
The integral from 0 to 4 of 3tsqrt(16-t^2) is 64