What is the integral from 0 to 4 of 8x ?

The integral from 0 to 4 of 8x is 64

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integral from 0 to 4 of 8x

\int_{0}^{4} 8 x d x

64

Solution steps

\int_{0}^{4} 8 x d x

Take the constant out:

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

= 8 \cdot \int_{0}^{4} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{4}

= 8 [\frac{x ^{2}}{2}]_{0}^{4}

Compute the boundaries:

8

= 8 \cdot 8

Simplify

= 64

\int \frac{x + 1}{x ^{2} + 2 x} d x

\frac{d y}{d x} = cos (2 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = 6 e^{t}

\frac{d}{d t} (x (t, y)), y = 4 x (t, y)

\frac{d}{d x} (\frac{1}{1 - x ^{3}})