What is the integral from 0 to 2 of sqrt(9x+37) ?

The integral from 0 to 2 of sqrt(9x+37) is (110sqrt(55)-74sqrt(37))/(27)

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integral from 0 to 2 of sqrt(9x+37)

\int_{0}^{2} 9 x + 37 d x

\frac{110 55 - 74 37}{27}

Decimal Notation

13.54286 \dots

Solution steps

\int_{0}^{2} 9 x + 37 d x

Apply u-substitution

: \int_{37}^{55} \frac{u}{9} d u

= \int_{37}^{55} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{37}^{55} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{37}^{55}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{37}^{55}

Compute the boundaries:

\frac{110 55 - 74 37}{3}

= \frac{1}{9} \cdot \frac{110 55 - 74 37}{3}

Simplify

= \frac{110 55 - 74 37}{27}

\frac{\partial}{\partial y} (x^{2 y})

\frac{d ^{2}}{d x ^{2}} (5 x^{4} - 5 x^{2} + 7)

y = 2 e^{x}

\int_{0}^{π} (9 e^{x} + 7 sin (x)) d x

\int x \cdot a^{x} d x

What is the integral from 0 to 2 of sqrt(9x+37) ?
The integral from 0 to 2 of sqrt(9x+37) is (110sqrt(55)-74sqrt(37))/(27)