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

The integral from 1 to 4 of 1/(x^3) is 15/32

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

\int_{1}^{4} \frac{1}{x ^{3}} d x

\frac{15}{32}

Decimal Notation

0.46875

Solution steps

\int_{1}^{4} \frac{1}{x ^{3}} d x

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{1}^{4}

= [- \frac{1}{2 x ^{2}}]_{1}^{4}

Compute the boundaries:

\frac{15}{32}

= \frac{15}{32}

\int_{0}^{2} \frac{1}{( 1 - x ) ^{3}} d x

\int_{- 1}^{0} (4 x - 1) d x

\int_{1}^{6} 5 + t d t

\int_{0}^{\frac{π}{2}} sin^{2} (u) d u

\int_{1}^{2} \frac{1}{x + x ^{2}} d x

What is the integral from 1 to 4 of 1/(x^3) ?
The integral from 1 to 4 of 1/(x^3) is 15/32