What is the integral from 0 to 1 of x^2sqrt(7x+8) ?

The integral from 0 to 1 of x^2sqrt(7x+8) is (17250sqrt(15)-16384sqrt(2))/(36015)

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integral from 0 to 1 of x^2sqrt(7x+8)

\int_{0}^{1} x^{2} 7 x + 8 d x

\frac{17250 15 - 16384 2}{36015}

Decimal Notation

1.21167 \dots

Solution steps

\int_{0}^{1} x^{2} 7 x + 8 d x

Apply u-substitution

: \int_{8}^{15} \frac{u ( u - 8 ) ^{2}}{343} d u

= \int_{8}^{15} \frac{u ( u - 8 ) ^{2}}{343} d u

Take the constant out:

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

= \frac{1}{343} \cdot \int_{8}^{15} u (u - 8)^{2} d u

Expand

u (u - 8)^{2} : u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 u

= \frac{1}{343} \cdot \int_{8}^{15} u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 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

= \frac{1}{343} (\int_{8}^{15} u^{\frac{5}{2}} d u - \int_{8}^{15} 16 u^{\frac{3}{2}} d u + \int_{8}^{15} 64 u d u)

\int_{8}^{15} u^{\frac{5}{2}} d u = \frac{6750 15 - 2048 2}{7}

\int_{8}^{15} 16 u^{\frac{3}{2}} d u = 16 (9015 - \frac{256 2}{5})

\int_{8}^{15} 64 u d u = 64 (1015 - \frac{32 2}{3})

= \frac{1}{343} (\frac{6750 15 - 2048 2}{7} - 16 (9015 - \frac{256 2}{5}) + 64 (1015 - \frac{32 2}{3}))

Simplify

\frac{1}{343} (\frac{6750 15 - 2048 2}{7} - 16 (9015 - \frac{256 2}{5}) + 64 (1015 - \frac{32 2}{3})) : \frac{17250 15 - 16384 2}{36015}

= \frac{17250 15 - 16384 2}{36015}

\int_{0}^{1} x^{2} (\frac{1}{2} + \frac{3}{2} x^{2}) d x

\int_{- 1}^{0} x^{2} 1 + x^{3} d x

\int_{- 2}^{2} ∣ x ∣ - ∣ x - 1 ∣ d x

\int_{0}^{2} (t i - t^{3} j + 3 t^{5} k) d t

\int_{1}^{3} x^{2} - x - 2 d x

What is the integral from 0 to 1 of x^2sqrt(7x+8) ?
The integral from 0 to 1 of x^2sqrt(7x+8) is (17250sqrt(15)-16384sqrt(2))/(36015)