Solve ∫ (from 1 to 3) of (4x^3-4x+1) wrt x

$$\int _ { 1 } ^ { 3 } ( 4 x ^ { 3 } - 4 x + 1 ) d x$$

Evaluate

$$66$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 -4 times \frac{x^{2}}{2}.

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

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$3^{4}-2\times 3^{2}+3-\left(1^{4}-2\times 1^{2}+1\right)$$

Simplify.

$$66$$