What is the integral from 5 to 8 of 2400xsqrt(3+2x) ?

The integral from 5 to 8 of 2400xsqrt(3+2x) is 63840sqrt(19)-24960sqrt(13)

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integral from 5 to 8 of 2400xsqrt(3+2x)

\int_{5}^{8} 2400 x 3 + 2 x d x

6384019 - 2496013

Decimal

188277.54872 \dots

Solution steps

\int_{5}^{8} 2400 x 3 + 2 x d x

Take the constant out:

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

= 2400 \cdot \int_{5}^{8} x 3 + 2 x d x

Apply u-substitution

: \int_{13}^{19} \frac{u ( u - 3 )}{4} d u

= 2400 \cdot \int_{13}^{19} \frac{u ( u - 3 )}{4} d u

Take the constant out:

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

= 2400 \cdot \frac{1}{4} \cdot \int_{13}^{19} u (u - 3) d u

Expand

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

= 2400 \cdot \frac{1}{4} \cdot \int_{13}^{19} u^{\frac{3}{2}} - 3 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

= 2400 \cdot \frac{1}{4} (\int_{13}^{19} u^{\frac{3}{2}} d u - \int_{13}^{19} 3 u d u)

\int_{13}^{19} u^{\frac{3}{2}} d u = \frac{722 19 - 338 13}{5}

\int_{13}^{19} 3 u d u = 3819 - 2613

= 2400 \cdot \frac{1}{4} (\frac{722 19 - 338 13}{5} - (3819 - 2613))

Simplify

2400 \cdot \frac{1}{4} (\frac{722 19 - 338 13}{5} - (3819 - 2613)) : 6384019 - 2496013

= 6384019 - 2496013

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What is the integral from 5 to 8 of 2400xsqrt(3+2x) ?
The integral from 5 to 8 of 2400xsqrt(3+2x) is 63840sqrt(19)-24960sqrt(13)