What is the integral from 4 to 6 of 700xsqrt(3+2x) ?

The integral from 4 to 6 of 700xsqrt(3+2x) is -4620sqrt(11)+10500sqrt(15)

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integral from 4 to 6 of 700xsqrt(3+2x)

\int_{4}^{6} 700 x 3 + 2 x d x

- 462011 + 1050015

Decimal

25343.51860 \dots

Solution steps

\int_{4}^{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_{4}^{6} x 3 + 2 x d x

Apply u-substitution

: \int_{11}^{15} \frac{u ( u - 3 )}{4} d u

= 700 \cdot \int_{11}^{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_{11}^{15} u (u - 3) d u

Expand

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

= 700 \cdot \frac{1}{4} \cdot \int_{11}^{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_{11}^{15} u^{\frac{3}{2}} d u - \int_{11}^{15} 3 u d u)

\int_{11}^{15} u^{\frac{3}{2}} d u = 9015 - \frac{242 11}{5}

\int_{11}^{15} 3 u d u = 3015 - 2211

= 700 \cdot \frac{1}{4} (9015 - \frac{242 11}{5} - (3015 - 2211))

Simplify

700 \cdot \frac{1}{4} (9015 - \frac{242 11}{5} - (3015 - 2211)) : - 462011 + 1050015

= - 462011 + 1050015

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What is the integral from 4 to 6 of 700xsqrt(3+2x) ?
The integral from 4 to 6 of 700xsqrt(3+2x) is -4620sqrt(11)+10500sqrt(15)