What is the integral from 1 to x of 8/(3t^3) ?

The integral from 1 to x of 8/(3t^3) is -4/(3x^2)+4/3

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integral from 1 to x of 8/(3t^3)

\int_{1}^{x} \frac{8}{3 t ^{3}} d t

- \frac{4}{3 x ^{2}} + \frac{4}{3}

Solution steps

\int_{1}^{x} \frac{8}{3 t ^{3}} d t

Take the constant out:

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

= \frac{8}{3} \cdot \int_{1}^{x} \frac{1}{t ^{3}} d t

Apply the Power Rule

: [- \frac{1}{2 t ^{2}}]_{1}^{x}

= \frac{8}{3} [- \frac{1}{2 t ^{2}}]_{1}^{x}

Compute the boundaries:

- \frac{1}{2 x ^{2}} + \frac{1}{2}

= \frac{8}{3} (- \frac{1}{2 x ^{2}} + \frac{1}{2})

Simplify

= - \frac{4}{3 x ^{2}} + \frac{4}{3}

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What is the integral from 1 to x of 8/(3t^3) ?
The integral from 1 to x of 8/(3t^3) is -4/(3x^2)+4/3