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

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

Evaluate

$$\frac{141}{10}=14.1$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}-\int x\mathrm{d}x+\int 2\mathrm{d}x$$

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

$$\frac{x^{5}}{5}-\frac{x^{2}}{2}+\int 2\mathrm{d}x$$

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}-\frac{x^{2}}{2}+2x$$

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.

$$\frac{1^{5}}{5}-\frac{1^{2}}{2}+2\times 1-\left(\frac{\left(-2\right)^{5}}{5}-\frac{\left(-2\right)^{2}}{2}+2\left(-2\right)\right)$$

Simplify.

$$\frac{141}{10}$$