What is the integral from 1 to 6 of 2/(x^3) ?

The integral from 1 to 6 of 2/(x^3) is 35/36

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integral from 1 to 6 of 2/(x^3)

\int_{1}^{6} \frac{2}{x ^{3}} d x

\frac{35}{36}

Decimal Notation

0.97222 \dots

Solution steps

\int_{1}^{6} \frac{2}{x ^{3}} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{6} \frac{1}{x ^{3}} d x

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{1}^{6}

= 2 [- \frac{1}{2 x ^{2}}]_{1}^{6}

Compute the boundaries:

\frac{35}{72}

= 2 \cdot \frac{35}{72}

Simplify

= \frac{35}{36}

6 - 4 x^{2}

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\int 4 x^{4} (27 + x^{5})^{\frac{1}{3}} d x

\int t sin (x) d x

\int_{- 2}^{3} (- x^{2} + x + 6) d x

What is the integral from 1 to 6 of 2/(x^3) ?
The integral from 1 to 6 of 2/(x^3) is 35/36