What is the integral from 0 to 4 of kx ?

The integral from 0 to 4 of kx is k*8

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integral from 0 to 4 of kx

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

k \cdot 8

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

8

= k \cdot 8

\int_{1}^{4} - 2 x + 8 d x

\int_{3}^{8} \frac{1}{1 + x - 1} d x

\int_{2}^{5} (x^{2} - 2 x + 3) d x

\int_{0}^{π} 1 - sin (2 x) d x

\int_{- 5}^{5} x (x^{12} - 7 x^{8} - 5) d x