Solve ∫ (from 0 to 1) of (4x-2) wrt x

$$\int _ { 0 } ^ { 1 } ( 4 x - 2 ) d x$$

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

$$2x^{2}-2x$$

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.

$$2\times 1^{2}-2-\left(2\times 0^{2}-2\times 0\right)$$

Simplify.

$$0$$