What is the integral from 0 to 2 of sqrt(36x+73) ?

The integral from 0 to 2 of sqrt(36x+73) is (145sqrt(145)-73sqrt(73))/(54)

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integral from 0 to 2 of sqrt(36x+73)

\int_{0}^{2} 36 x + 73 d x

\frac{145 145 - 73 73}{54}

Decimal

20.78368 \dots

Solution steps

\int_{0}^{2} 36 x + 73 d x

Apply u-substitution

: \int_{73}^{145} \frac{u}{36} d u

= \int_{73}^{145} \frac{u}{36} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{36} \cdot \int_{73}^{145} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{73}^{145}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{73}^{145}

Compute the boundaries:

\frac{290 145 - 146 73}{3}

= \frac{1}{36} \cdot \frac{290 145 - 146 73}{3}

Simplify

= \frac{145 145 - 73 73}{54}

\frac{\partial}{\partial x} (I n (3 x^{2} y))

x \to 0 + lim (3 - x)

f (x) = \frac{r}{3}

\int 1 0^{u} d u

\frac{\partial}{\partial x} (2 x^{2} - y^{2} + 3 x - y)

What is the integral from 0 to 2 of sqrt(36x+73) ?
The integral from 0 to 2 of sqrt(36x+73) is (145sqrt(145)-73sqrt(73))/(54)