Solve ∫ (from 1 to 2) of x^2+1/x^4 wrt x=

$$\int _ { 1 } ^ { 2 } x ^ { 2 } + \frac { 1 } { x ^ { 4 } } d x =$$

Evaluate

$$\frac{21}{8}=2.625$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+\frac{1}{x^{4}}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int \frac{1}{x^{4}}\mathrm{d}x$$

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

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

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

$$\frac{x^{3}}{3}-\frac{1}{3x^{3}}$$

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.

$$\frac{2^{3}}{3}-\frac{2^{-3}}{3}-\left(\frac{1^{3}}{3}-\frac{1^{-3}}{3}\right)$$

Simplify.

$$\frac{21}{8}$$