What is the integral from 0 to 1 of 6tsqrt(4t^2+1) ?

The integral from 0 to 1 of 6tsqrt(4t^2+1) is (5sqrt(5)-1)/2

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integral from 0 to 1 of 6tsqrt(4t^2+1)

\int_{0}^{1} 6 t 4 t^{2} + 1 d t

\frac{5 5 - 1}{2}

Decimal

5.09016 \dots

Solution steps

\int_{0}^{1} 6 t 4 t^{2} + 1 d t

Take the constant out:

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

= 6 \cdot \int_{0}^{1} t 4 t^{2} + 1 d t

Apply u-substitution

: \int_{1}^{5} \frac{u}{8} d u

= 6 \cdot \int_{1}^{5} \frac{u}{8} d u

Take the constant out:

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

= 6 \cdot \frac{1}{8} \cdot \int_{1}^{5} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= 6 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Simplify

6 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5} : \frac{3}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= \frac{3}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Compute the boundaries:

\frac{10 5}{3} - \frac{2}{3}

= \frac{3}{4} (\frac{10 5}{3} - \frac{2}{3})

Simplify

= \frac{5 5 - 1}{2}

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What is the integral from 0 to 1 of 6tsqrt(4t^2+1) ?
The integral from 0 to 1 of 6tsqrt(4t^2+1) is (5sqrt(5)-1)/2