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

The integral from 0 to 2 of 2t(sqrt(t^2+1)) is (2(5sqrt(5)-1))/3

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

\int_{0}^{2} 2 t (t^{2} + 1) d t

\frac{2 ( 5 5 - 1 )}{3}

Decimal Notation

6.78689 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{5} u^{2} d u

= 2 \cdot \int_{1}^{5} u^{2} d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{5 5}{3} - \frac{1}{3}

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

Simplify

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

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What is the integral from 0 to 2 of 2t(sqrt(t^2+1)) ?
The integral from 0 to 2 of 2t(sqrt(t^2+1)) is (2(5sqrt(5)-1))/3