Solve ∫ (from 0 to 1) of 2x^3-6x^2+4x wrt x

$$\int_{ 0 }^{ 1 } 2 { x }^{ 3 } -6 { x }^{ 2 } +4x d x$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x^{3}-6x^{2}+4x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{3}\mathrm{d}x+\int -6x^{2}\mathrm{d}x+\int 4x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{3}\mathrm{d}x-6\int x^{2}\mathrm{d}x+4\int x\mathrm{d}x$$

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

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

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}.

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

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}.

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

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.

$$\frac{1^{4}}{2}-2\times 1^{3}+2\times 1^{2}-\left(\frac{0^{4}}{2}-2\times 0^{3}+2\times 0^{2}\right)$$

Simplify.

$$\frac{1}{2}$$