What is the integral from 1 to 2 of tsqrt(1/2 t^2+9) ?

The integral from 1 to 2 of tsqrt(1/2 t^2+9) is (22sqrt(11))/3-(19sqrt(19))/(3sqrt(2))

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integral from 1 to 2 of tsqrt(1/2 t^2+9)

\int_{1}^{2} t \frac{1}{2} t^{2} + 9 d t

\frac{22 11}{3} - \frac{19 19}{3 2}

Decimal

4.80127 \dots

Solution steps

\int_{1}^{2} t \frac{1}{2} t^{2} + 9 d t

Apply Integration By Parts

: [\frac{1}{2 2} t^{2} t^{2} + 18 - \int \frac{t ^{3}}{2 2 t ^{2} + 18} d t]_{1}^{2}

= [\frac{1}{2 2} t^{2} t^{2} + 18 - \int \frac{t ^{3}}{2 2 t ^{2} + 18} d t]_{1}^{2}

\int \frac{t ^{3}}{2 2 t ^{2} + 18} d t = \frac{1}{2 2} (\frac{1}{3} (t^{2} + 18)^{\frac{3}{2}} - 18 t^{2} + 18)

= [\frac{1}{2 2} t^{2} t^{2} + 18 - \frac{1}{2 2} (\frac{1}{3} (t^{2} + 18)^{\frac{3}{2}} - 18 t^{2} + 18)]_{1}^{2}

Simplify

[\frac{1}{2 2} t^{2} t^{2} + 18 - \frac{1}{2 2} (\frac{1}{3} (t^{2} + 18)^{\frac{3}{2}} - 18 t^{2} + 18)]_{1}^{2} : [\frac{1}{3 2} (t^{2} + 18) t^{2} + 18]_{1}^{2}

= [\frac{1}{3 2} (t^{2} + 18) t^{2} + 18]_{1}^{2}

Compute the boundaries:

\frac{22 11}{3} - \frac{19 19}{3 2}

= \frac{22 11}{3} - \frac{19 19}{3 2}

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What is the integral from 1 to 2 of tsqrt(1/2 t^2+9) ?
The integral from 1 to 2 of tsqrt(1/2 t^2+9) is (22sqrt(11))/3-(19sqrt(19))/(3sqrt(2))