What is the integral from 0 to 1 of 2t^3sqrt(4t^2+4) ?

The integral from 0 to 1 of 2t^3sqrt(4t^2+4) is 2*(4sqrt(2)+4)/(15)

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integral from 0 to 1 of 2t^3sqrt(4t^2+4)

\int_{0}^{1} 2 t^{3} 4 t^{2} + 4 d t

2 \cdot \frac{4 2 + 4}{15}

Decimal

1.28758 \dots

Solution steps

\int_{0}^{1} 2 t^{3} 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}^{1} t^{3} 4 t^{2} + 4 d t

4 t^{2} + 4 = 2 t^{2} + 1

= 2 \cdot \int_{0}^{1} 2 t^{3} t^{2} + 1 d t

Take the constant out:

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

= 2 \cdot 2 \cdot \int_{0}^{1} t^{3} t^{2} + 1 d t

Apply u-substitution

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

Take the constant out:

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

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

Expand

(u - 1) u : u^{\frac{3}{2}} - u

= 2 \cdot 2 \cdot \frac{1}{2} \cdot \int_{1}^{2} u^{\frac{3}{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

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

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

\int_{1}^{2} u d u = \frac{4 2 - 2}{3}

= 2 \cdot 2 \cdot \frac{1}{2} (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3})

Simplify

2 \cdot 2 \cdot \frac{1}{2} (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3}) : 2 \cdot \frac{4 2 + 4}{15}

= 2 \cdot \frac{4 2 + 4}{15}

What is the integral from 0 to 1 of 2t^3sqrt(4t^2+4) ?
The integral from 0 to 1 of 2t^3sqrt(4t^2+4) is 2*(4sqrt(2)+4)/(15)