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

The integral from 0 to 4 of 2tsqrt(1+t^2) is (2(17sqrt(17)-1))/3

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

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

\frac{2 ( 17 17 - 1 )}{3}

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{17} u^{2} d u

= 2 \cdot \int_{1}^{17} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{1}^{17}

= 2 [\frac{u ^{3}}{3}]_{1}^{17}

Compute the boundaries:

\frac{17 17}{3} - \frac{1}{3}

= 2 (\frac{17 17}{3} - \frac{1}{3})

Simplify

= \frac{2 ( 17 17 - 1 )}{3}

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What is the integral from 0 to 4 of 2tsqrt(1+t^2) ?
The integral from 0 to 4 of 2tsqrt(1+t^2) is (2(17sqrt(17)-1))/3