What is the integral from x-2 to x^2 of sqrt(t-1) ?

The integral from x-2 to x^2 of sqrt(t-1) is (2(x^2-1)^{3/2}-2(x-3)^{3/2})/3

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integral from x-2 to x^2 of sqrt(t-1)

\int_{x - 2}^{x^{2}} t - 1 d t

\frac{2 ( x ^{2} - 1 ) ^{\frac{3}{2}} - 2 ( x - 3 ) ^{\frac{3}{2}}}{3}

Solution steps

\int_{x - 2}^{x^{2}} t - 1 d t

Apply u-substitution

: \int_{x - 3}^{x^{2} - 1} u d u

= \int_{x - 3}^{x^{2} - 1} u d u

Apply the Power Rule

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

= [\frac{2}{3} u^{\frac{3}{2}}]_{x - 3}^{x^{2} - 1}

Compute the boundaries:

\frac{2}{3} (x^{2} - 1)^{\frac{3}{2}} - \frac{2}{3} (x - 3)^{\frac{3}{2}}

= \frac{2}{3} (x^{2} - 1)^{\frac{3}{2}} - \frac{2}{3} (x - 3)^{\frac{3}{2}}

Simplify

= \frac{2 ( x ^{2} - 1 ) ^{\frac{3}{2}} - 2 ( x - 3 ) ^{\frac{3}{2}}}{3}

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What is the integral from x-2 to x^2 of sqrt(t-1) ?
The integral from x-2 to x^2 of sqrt(t-1) is (2(x^2-1)^{3/2}-2(x-3)^{3/2})/3