What is the integral from 1 to 8 of sqrt(1+3x) ?

The integral from 1 to 8 of sqrt(1+3x) is 26

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integral from 1 to 8 of sqrt(1+3x)

\int_{1}^{8} 1 + 3 x d x

26

Solution steps

\int_{1}^{8} 1 + 3 x d x

Apply u-substitution

= \int_{4}^{25} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{25} u d u

Apply the Power Rule

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{25}

Compute the boundaries:

78

= \frac{1}{3} \cdot 78

Simplify

= 26

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\frac{\partial}{\partial x} (x cos (y - x))

\int_{- 3}^{0} \frac{1}{9 - x ^{2}} d x

What is the integral from 1 to 8 of sqrt(1+3x) ?
The integral from 1 to 8 of sqrt(1+3x) is 26