What is the integral from 0 to 1 of tsqrt(1+36t^2) ?

The integral from 0 to 1 of tsqrt(1+36t^2) is (37sqrt(37)-1)/(108)

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integral from 0 to 1 of tsqrt(1+36t^2)

\int_{0}^{1} t 1 + 36 t^{2} d t

\frac{37 37 - 1}{108}

Decimal Notation

2.07465 \dots

Solution steps

\int_{0}^{1} t 1 + 36 t^{2} d t

Apply u-substitution

: \int_{1}^{37} \frac{u}{72} d u

= \int_{1}^{37} \frac{u}{72} d u

Take the constant out:

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

= \frac{1}{72} \cdot \int_{1}^{37} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

= \frac{1}{72} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

Compute the boundaries:

\frac{74 37}{3} - \frac{2}{3}

= \frac{1}{72} (\frac{74 37}{3} - \frac{2}{3})

Simplify

= \frac{37 37 - 1}{108}

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What is the integral from 0 to 1 of tsqrt(1+36t^2) ?
The integral from 0 to 1 of tsqrt(1+36t^2) is (37sqrt(37)-1)/(108)