Solve ∫ (from - 4 to - 1) of -41/x^3 wrt x

$$\int _ { - 4 } ^ { - 1 } - 4 \frac { 1 } { x ^ { 3 } } d x$$

Evaluate

$$\frac{15}{8}=1.875$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -\frac{4}{x^{3}}\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$-4\int \frac{1}{x^{3}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{3}}\mathrm{d}x with -\frac{1}{2x^{2}}. Multiply -4 times -\frac{1}{2x^{2}}.

$$\frac{2}{x^{2}}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$2\left(-1\right)^{-2}-2\left(-4\right)^{-2}$$

Simplify.

$$\frac{15}{8}$$