What is the integral from 1 to 4 of sqrt(x-1) ?

The integral from 1 to 4 of sqrt(x-1) is 2sqrt(3)

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integral from 1 to 4 of sqrt(x-1)

\int_{1}^{4} x - 1 d x

23

Decimal

3.46410 \dots

Solution steps

\int_{1}^{4} x - 1 d x

Apply u-substitution

: \int_{0}^{3} u d u

= \int_{0}^{3} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{3}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{3}

Compute the boundaries:

23

= 23

\int_{0}^{x^{2}} \frac{1}{3 + x ^{3}} d y

\int_{0}^{7} 8 d x

\int_{\frac{π}{2}}^{π} x cos (x) d x

\int_{1}^{3} \frac{1}{y} d y

\int_{0}^{1} x arctan^{2} (x) d x

What is the integral from 1 to 4 of sqrt(x-1) ?
The integral from 1 to 4 of sqrt(x-1) is 2sqrt(3)