What is the integral from 1 to 3 of 2/(t^3) ?

The integral from 1 to 3 of 2/(t^3) is 8/9

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integral from 1 to 3 of 2/(t^3)

\int_{1}^{3} \frac{2}{t ^{3}} d t

\frac{8}{9}

Decimal Notation

0.88888 \dots

Solution steps

\int_{1}^{3} \frac{2}{t ^{3}} d t

Take the constant out:

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

= 2 \cdot \int_{1}^{3} \frac{1}{t ^{3}} d t

Apply the Power Rule

: [- \frac{1}{2 t ^{2}}]_{1}^{3}

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

Compute the boundaries:

\frac{4}{9}

= 2 \cdot \frac{4}{9}

Simplify

= \frac{8}{9}

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

\int a - x^{2} d x

n = 1 \sum \infty \frac{1}{3 n - 2}

f (x) = 6 x^{l n (x)}

- \frac{5}{x ^{3}}

What is the integral from 1 to 3 of 2/(t^3) ?
The integral from 1 to 3 of 2/(t^3) is 8/9