What is the integral from 6 to 9 of 2400xsqrt(3+2x) ?

The integral from 6 to 9 of 2400xsqrt(3+2x) is 80640sqrt(21)-36000sqrt(15)

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integral from 6 to 9 of 2400xsqrt(3+2x)

\int_{6}^{9} 2400 x 3 + 2 x d x

8064021 - 3600015

Decimal Notation

230111.50357 \dots

Solution steps

\int_{6}^{9} 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_{6}^{9} x 3 + 2 x d x

Apply u-substitution

: \int_{15}^{21} \frac{u ( u - 3 )}{4} d u

= 2400 \cdot \int_{15}^{21} \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_{15}^{21} u (u - 3) d u

Expand

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

= 2400 \cdot \frac{1}{4} \cdot \int_{15}^{21} 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_{15}^{21} u^{\frac{3}{2}} d u - \int_{15}^{21} 3 u d u)

\int_{15}^{21} u^{\frac{3}{2}} d u = \frac{882 21}{5} - 9015

\int_{15}^{21} 3 u d u = 3 (1421 - 1015)

= 2400 \cdot \frac{1}{4} (\frac{882 21}{5} - 9015 - 3 (1421 - 1015))

Simplify

2400 \cdot \frac{1}{4} (\frac{882 21}{5} - 9015 - 3 (1421 - 1015)) : 8064021 - 3600015

= 8064021 - 3600015

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What is the integral from 6 to 9 of 2400xsqrt(3+2x) ?
The integral from 6 to 9 of 2400xsqrt(3+2x) is 80640sqrt(21)-36000sqrt(15)