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

The integral from 0 to 2 of 3xsqrt(x^2+4a^2) is 8((a^2+1)^{3/2}-a^3)

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

\int_{0}^{2} 3 x x^{2} + 4 a^{2} d x

8 ((a^{2} + 1)^{\frac{3}{2}} - a^{3})

Solution steps

\int_{0}^{2} 3 x x^{2} + 4 a^{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 x^{2} + 4 a^{2} d x

Apply u-substitution

: \int_{4 a^{2}}^{4 + 4 a^{2}} \frac{u}{2} d u

= 3 \cdot \int_{4 a^{2}}^{4 + 4 a^{2}} \frac{u}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{4 a^{2}}^{4 + 4 a^{2}} u d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{4 a^{2}}^{4 + 4 a^{2}}

Simplify

3 \cdot \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{4 a^{2}}^{4 + 4 a^{2}} : \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{4 a^{2}}^{4 + 4 a^{2}}

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

Compute the boundaries:

\frac{16}{3} (1 + a^{2})^{\frac{3}{2}} - \frac{16}{3} a^{3}

= \frac{3}{2} (\frac{16}{3} (1 + a^{2})^{\frac{3}{2}} - \frac{16}{3} a^{3})

Simplify

= 8 ((a^{2} + 1)^{\frac{3}{2}} - a^{3})

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What is the integral from 0 to 2 of 3xsqrt(x^2+4a^2) ?
The integral from 0 to 2 of 3xsqrt(x^2+4a^2) is 8((a^2+1)^{3/2}-a^3)