What is the integral from 2 to 3 of tsqrt(1+4t^2) ?

The integral from 2 to 3 of tsqrt(1+4t^2) is (37sqrt(37)-17sqrt(17))/(12)

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integral from 2 to 3 of tsqrt(1+4t^2)

\int_{2}^{3} t 1 + 4 t^{2} d t

\frac{37 37 - 17 17}{12}

Decimal Notation

12.91411 \dots

Solution steps

\int_{2}^{3} t 1 + 4 t^{2} d t

Apply u-substitution

: \int_{17}^{37} \frac{u}{8} d u

= \int_{17}^{37} \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{17}^{37} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{17}^{37}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{17}^{37}

Compute the boundaries:

\frac{74 37 - 34 17}{3}

= \frac{1}{8} \cdot \frac{74 37 - 34 17}{3}

Simplify

= \frac{37 37 - 17 17}{12}

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What is the integral from 2 to 3 of tsqrt(1+4t^2) ?
The integral from 2 to 3 of tsqrt(1+4t^2) is (37sqrt(37)-17sqrt(17))/(12)