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

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

Evaluate

$$\frac{129}{2}=64.5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

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

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

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.

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

Simplify.

$$\frac{129}{2}$$