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

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

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

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

55 - 1

Decimal

10.18033 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \int_{1}^{5} u^{2} d u

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

= 55 - 1

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