What is the integral from 1 to 2 of tsqrt(1/2 t^2+6) ?

The integral from 1 to 2 of tsqrt(1/2 t^2+6) is (32sqrt(2))/3-(13sqrt(13))/(3sqrt(2))

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integral from 1 to 2 of tsqrt(1/2 t^2+6)

\int_{1}^{2} t \frac{1}{2} t^{2} + 6 d t

\frac{32 2}{3} - \frac{13 13}{3 2}

Decimal

4.03706 \dots

Solution steps

\int_{1}^{2} t \frac{1}{2} t^{2} + 6 d t

Apply Integration By Parts

: [\frac{1}{2 2} t^{2} t^{2} + 12 - \int \frac{t ^{3}}{2 2 t ^{2} + 12} d t]_{1}^{2}

= [\frac{1}{2 2} t^{2} t^{2} + 12 - \int \frac{t ^{3}}{2 2 t ^{2} + 12} d t]_{1}^{2}

\int \frac{t ^{3}}{2 2 t ^{2} + 12} d t = \frac{1}{2 2} (\frac{1}{3} (t^{2} + 12)^{\frac{3}{2}} - 12 t^{2} + 12)

= [\frac{1}{2 2} t^{2} t^{2} + 12 - \frac{1}{2 2} (\frac{1}{3} (t^{2} + 12)^{\frac{3}{2}} - 12 t^{2} + 12)]_{1}^{2}

Simplify

[\frac{1}{2 2} t^{2} t^{2} + 12 - \frac{1}{2 2} (\frac{1}{3} (t^{2} + 12)^{\frac{3}{2}} - 12 t^{2} + 12)]_{1}^{2} : [\frac{1}{3 2} (t^{2} + 12) t^{2} + 12]_{1}^{2}

= [\frac{1}{3 2} (t^{2} + 12) t^{2} + 12]_{1}^{2}

Compute the boundaries:

\frac{32 2}{3} - \frac{13 13}{3 2}

= \frac{32 2}{3} - \frac{13 13}{3 2}

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What is the integral from 1 to 2 of tsqrt(1/2 t^2+6) ?
The integral from 1 to 2 of tsqrt(1/2 t^2+6) is (32sqrt(2))/3-(13sqrt(13))/(3sqrt(2))