What is the integral from 0 to 2 of 3xsqrt(4-x^2) ?

The integral from 0 to 2 of 3xsqrt(4-x^2) is 8

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integral from 0 to 2 of 3xsqrt(4-x^2)

\int_{0}^{2} 3 x 4 - x^{2} d x

8

Solution steps

\int_{0}^{2} 3 x 4 - x^{2} d x

Take the constant out:

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

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

Apply u-substitution

: \int_{4}^{0} - \frac{u}{2} d u

= 3 \cdot \int_{4}^{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}^{4} - \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}^{4} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= 3 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}))

Simplify

3 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4})) : \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{3}{2} \cdot \frac{16}{3}

Simplify

= 8

\int_{- 11}^{1} x 5 - x d x

\int_{0}^{2} - x^{4} + 2 x^{2} + 8 d x

\int_{5}^{10} \frac{18}{25} x d x

\int_{- 2}^{1} (3 x^{2} + 2) d x

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

What is the integral from 0 to 2 of 3xsqrt(4-x^2) ?
The integral from 0 to 2 of 3xsqrt(4-x^2) is 8