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

The integral from 0 to 3 of 2tsqrt(4t^2+4) is (4(10sqrt(10)-1))/3

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

\int_{0}^{3} 2 t 4 t^{2} + 4 d t

\frac{4 ( 10 10 - 1 )}{3}

Solution steps

\int_{0}^{3} 2 t 4 t^{2} + 4 d t

Take the constant out:

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

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

Apply u-substitution

: \int_{4}^{40} \frac{u}{8} d u

= 2 \cdot \int_{4}^{40} \frac{u}{8} d u

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int_{4}^{40} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{40}

= 2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{40}

Simplify

2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{40} : \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{40}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{40}

Compute the boundaries:

\frac{160 10}{3} - \frac{16}{3}

= \frac{1}{4} (\frac{160 10}{3} - \frac{16}{3})

Simplify

= \frac{4 ( 10 10 - 1 )}{3}

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What is the integral from 0 to 3 of 2tsqrt(4t^2+4) ?
The integral from 0 to 3 of 2tsqrt(4t^2+4) is (4(10sqrt(10)-1))/3