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

The integral from 1 to 2t of usqrt(9+u^2) is ((9+4t^2)^{3/2}-10sqrt(10))/3

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integral from 1 to 2t of usqrt(9+u^2)

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

\frac{( 9 + 4 t ^{2} ) ^{\frac{3}{2}} - 10 10}{3}

Solution steps

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

Apply u-substitution

: \int_{10}^{9 + 4 t^{2}} v^{2} d v

= \int_{10}^{9 + 4 t^{2}} v^{2} d v

Apply the Power Rule

: [\frac{v ^{3}}{3}]_{10}^{9 + 4 t^{2}}

= [\frac{v ^{3}}{3}]_{10}^{9 + 4 t^{2}}

Compute the boundaries:

\frac{( 9 + 4 t ^{2} ) ^{\frac{3}{2}}}{3} - \frac{10 10}{3}

= \frac{( 9 + 4 t ^{2} ) ^{\frac{3}{2}}}{3} - \frac{10 10}{3}

Simplify

= \frac{( 9 + 4 t ^{2} ) ^{\frac{3}{2}} - 10 10}{3}

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What is the integral from 1 to 2t of usqrt(9+u^2) ?
The integral from 1 to 2t of usqrt(9+u^2) is ((9+4t^2)^{3/2}-10sqrt(10))/3