What is the integral from 3 to 6 of 700xsqrt(3+2x) ?

The integral from 3 to 6 of 700xsqrt(3+2x) is 175(60sqrt(15)-216/5)

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integral from 3 to 6 of 700xsqrt(3+2x)

\int_{3}^{6} 700 x 3 + 2 x d x

175 (6015 - \frac{216}{5})

Decimal

33106.32513 \dots

Solution steps

\int_{3}^{6} 700 x 3 + 2 x d x

Take the constant out:

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

= 700 \cdot \int_{3}^{6} x 3 + 2 x d x

Apply u-substitution

: \int_{9}^{15} \frac{u ( u - 3 )}{4} d u

= 700 \cdot \int_{9}^{15} \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

= 700 \cdot \frac{1}{4} \cdot \int_{9}^{15} u (u - 3) d u

Expand

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

= 700 \cdot \frac{1}{4} \cdot \int_{9}^{15} 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

= 700 \cdot \frac{1}{4} (\int_{9}^{15} u^{\frac{3}{2}} d u - \int_{9}^{15} 3 u d u)

\int_{9}^{15} u^{\frac{3}{2}} d u = 9015 - \frac{486}{5}

\int_{9}^{15} 3 u d u = 3 (1015 - 18)

= 700 \cdot \frac{1}{4} (9015 - \frac{486}{5} - 3 (1015 - 18))

Simplify

700 \cdot \frac{1}{4} (9015 - \frac{486}{5} - 3 (1015 - 18)) : 175 (6015 - \frac{216}{5})

= 175 (6015 - \frac{216}{5})

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What is the integral from 3 to 6 of 700xsqrt(3+2x) ?
The integral from 3 to 6 of 700xsqrt(3+2x) is 175(60sqrt(15)-216/5)