What is the integral from 0 to 1 of tsqrt(1+16t^2) ?

The integral from 0 to 1 of tsqrt(1+16t^2) is (17sqrt(17)-1)/(48)

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integral from 0 to 1 of tsqrt(1+16t^2)

\int_{0}^{1} t 1 + 16 t^{2} d t

\frac{17 17 - 1}{48}

Decimal

1.43943 \dots

Solution steps

\int_{0}^{1} t 1 + 16 t^{2} d t

Apply u-substitution

: \int_{1}^{17} \frac{u}{32} d u

= \int_{1}^{17} \frac{u}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int_{1}^{17} u d u

Apply the Power Rule

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

= \frac{1}{32} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Compute the boundaries:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{32} (\frac{34 17}{3} - \frac{2}{3})

Simplify

= \frac{17 17 - 1}{48}

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What is the integral from 0 to 1 of tsqrt(1+16t^2) ?
The integral from 0 to 1 of tsqrt(1+16t^2) is (17sqrt(17)-1)/(48)