What is the integral from-infinity to-4 of 7/(x^3) ?

The integral from-infinity to-4 of 7/(x^3) is -7/32

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integral from-infinity to-4 of 7/(x^3)

\int_{- \infty}^{- 4} \frac{7}{x ^{3}} d x

- \frac{7}{32}

Decimal

- 0.21875

Solution steps

\int_{- \infty}^{- 4} \frac{7}{x ^{3}} d x

Compute the indefinite integral:

\int \frac{7}{x ^{3}} d x = - \frac{7}{2 x ^{2}} + C

Compute the boundaries:

\int_{- \infty}^{- 4} \frac{7}{x ^{3}} d x = - \frac{7}{32} - 0

= - \frac{7}{32} - 0

Simplify

= - \frac{7}{32}

\int_{0}^{1} x 99 x^{2} + 1 d x

\int_{- 2}^{1} 4 - x^{2} - x - 2 d x

\int_{1}^{4} (x^{2} + 5 x - 12) d x

\int_{- 4}^{6} 12 + x - \frac{1}{2} x^{2} d x

\int_{0}^{1} n x^{n} d x

What is the integral from-infinity to-4 of 7/(x^3) ?
The integral from-infinity to-4 of 7/(x^3) is -7/32