Solve ∫ (from 2 to 3) of (x^4-6x^2-2x+4) wrt x

$$\int _ { 2 } ^ { 3 } ( x ^ { 4 } - 6 x ^ { 2 } - 2 x + 4 ) d x$$

Evaluate

$$\frac{16}{5}=3.2$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{5}}{5}-6\int x^{2}\mathrm{d}x-2\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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}. Multiply -6 times \frac{x^{3}}{3}.

$$\frac{x^{5}}{5}-2x^{3}-2\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-2x^{3}-x^{2}+\int 4\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}-2x^{3}-x^{2}+4x$$

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.

$$\frac{3^{5}}{5}-2\times 3^{3}-3^{2}+4\times 3-\left(\frac{2^{5}}{5}-2\times 2^{3}-2^{2}+4\times 2\right)$$

Simplify.

$$\frac{16}{5}$$