What is the integral from 0 to 1 of 5xsqrt(2x+9) ?

The integral from 0 to 1 of 5xsqrt(2x+9) is 81-22sqrt(11)

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integral from 0 to 1 of 5xsqrt(2x+9)

\int_{0}^{1} 5 x 2 x + 9 d x

81 - 2211

Decimal

8.03425 \dots

Solution steps

\int_{0}^{1} 5 x 2 x + 9 d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} x 2 x + 9 d x

Apply u-substitution

= 5 \cdot \int_{9}^{11} \frac{u ( u - 9 )}{4} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int_{9}^{11} u (u - 9) d u

Expand

u (u - 9) : u^{\frac{3}{2}} - 9 u

= 5 \cdot \frac{1}{4} \cdot \int_{9}^{11} u^{\frac{3}{2}} - 9 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 \cdot \frac{1}{4} (\int_{9}^{11} u^{\frac{3}{2}} d u - \int_{9}^{11} 9 u d u)

\int_{9}^{11} u^{\frac{3}{2}} d u = \frac{242 11 - 486}{5}

\int_{9}^{11} 9 u d u = 9 (\frac{22 11}{3} - 18)

= 5 \cdot \frac{1}{4} (\frac{242 11 - 486}{5} - 9 (\frac{22 11}{3} - 18))

Simplify

5 \cdot \frac{1}{4} (\frac{242 11 - 486}{5} - 9 (\frac{22 11}{3} - 18)) : 81 - 2211

= 81 - 2211

What is the integral from 0 to 1 of 5xsqrt(2x+9) ?
The integral from 0 to 1 of 5xsqrt(2x+9) is 81-22sqrt(11)