What is the integral from 60 to x of 1/(-0.4x^3) ?

The integral from 60 to x of 1/(-0.4x^3) is -2.5(-(0.5)/(x^2)+0.00013…)

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integral from 60 to x of 1/(-0.4x^3)

\int_{60}^{x} \frac{1}{- 0.4 x ^{3}} d x

- 2.5 (- \frac{0.5}{x ^{2}} + 0.00013 \dots)

Solution steps

\int_{60}^{x} \frac{1}{- 0.4 x ^{3}} d x

Take the constant out:

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

= \frac{1}{- 0.4} \cdot \int_{60}^{x} \frac{1}{x ^{3}} d x

Simplify

= (- 2.5) \cdot \int_{60}^{x} \frac{1}{x ^{3}} d x

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{60}^{x}

= (- 2.5) [- \frac{1}{2 x ^{2}}]_{60}^{x}

Simplify

= - 2.5 [- \frac{0.5}{x ^{2}}]_{60}^{x}

Compute the boundaries:

- \frac{0.5}{x ^{2}} + 0.00013 \dots

= - 2.5 (- \frac{0.5}{x ^{2}} + 0.00013 \dots)

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What is the integral from 60 to x of 1/(-0.4x^3) ?
The integral from 60 to x of 1/(-0.4x^3) is -2.5(-(0.5)/(x^2)+0.00013…)