Solve ∫ (from - 2 to 1) of 1/2x^4-9/2x^2-2x+6 wrt x

$$\int _ { - 2 } ^ { 1 } \frac { 1 } { 2 } x ^ { 4 } - \frac { 9 } { 2 } x ^ { 2 } - 2 x + 6 d x$$

Evaluate

$$\frac{54}{5}=10.8$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{x^{4}}{2}-\frac{9x^{2}}{2}-2x+6\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{x^{4}}{2}\mathrm{d}x+\int -\frac{9x^{2}}{2}\mathrm{d}x+\int -2x\mathrm{d}x+\int 6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{\int x^{4}\mathrm{d}x}{2}-\frac{9\int x^{2}\mathrm{d}x}{2}-2\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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}. Multiply \frac{1}{2} times \frac{x^{5}}{5}.

$$\frac{x^{5}}{10}-\frac{9\int x^{2}\mathrm{d}x}{2}-2\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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}. Multiply -\frac{9}{2} times \frac{x^{3}}{3}.

$$\frac{x^{5}}{10}-\frac{3x^{3}}{2}-2\int x\mathrm{d}x+\int 6\mathrm{d}x$$

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 -2 times \frac{x^{2}}{2}.

$$\frac{x^{5}}{10}-\frac{3x^{3}}{2}-x^{2}+\int 6\mathrm{d}x$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{10}-\frac{3x^{3}}{2}-x^{2}+6x$$

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.

$$\frac{1^{5}}{10}-\frac{3}{2}\times 1^{3}-1^{2}+6\times 1-\left(\frac{\left(-2\right)^{5}}{10}-\frac{3}{2}\left(-2\right)^{3}-\left(-2\right)^{2}+6\left(-2\right)\right)$$

Simplify.

$$\frac{54}{5}$$