What is the integral of 7x(8-x) ?

The integral of 7x(8-x) is 7(4x^2-(x^3)/3)+C

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integral of 7x(8-x)

\int 7 x (8 - x) d x

7 (4 x^{2} - \frac{x ^{3}}{3}) + C

Solution steps

\int 7 x (8 - x) d x

Take the constant out:

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

= 7 \cdot \int x (8 - x) d x

Expand

x (8 - x) : 8 x - x^{2}

= 7 \cdot \int 8 x - x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 (\int 8 x d x - \int x^{2} d x)

\int 8 x d x = 4 x^{2}

\int x^{2} d x = \frac{x ^{3}}{3}

= 7 (4 x^{2} - \frac{x ^{3}}{3})

Add a constant to the solution

= 7 (4 x^{2} - \frac{x ^{3}}{3}) + C

y^{2} = 4 + x, y^{2} + x = 2

f (x) = (1 + 2 x^{2})^{2} + 2 x^{3}

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

\frac{d}{d x} (arcsin (x) + x 1 - x^{2})

\frac{d}{d x} (4 x^{2} + 4 x^{3})