What is the integral from 1 to 3 of 9/(x^3) ?

The integral from 1 to 3 of 9/(x^3) is 4

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integral from 1 to 3 of 9/(x^3)

\int_{1}^{3} \frac{9}{x ^{3}} d x

4

Solution steps

\int_{1}^{3} \frac{9}{x ^{3}} d x

Take the constant out:

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

= 9 \cdot \int_{1}^{3} \frac{1}{x ^{3}} d x

Apply the Power Rule

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

= 9 [- \frac{1}{2 x ^{2}}]_{1}^{3}

Compute the boundaries:

\frac{4}{9}

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

Simplify

= 4

\int_{0}^{1} arctan (d) x d x

\int_{0}^{1} \frac{5}{1 + p ^{2}} d p

\int_{- 5}^{6} \frac{1}{2} x d x

\int_{0}^{1} 5 z^{4} d z

\int_{3}^{0} 15 w^{4} - 13 w^{2} + w d w

What is the integral from 1 to 3 of 9/(x^3) ?
The integral from 1 to 3 of 9/(x^3) is 4