Solve ∫ (from 0 to 1) of (x^3-6x^2+5x) wrt x

$$\int _ { 0 } ^ { 1 } ( x ^ { 3 } - 6 x ^ { 2 } + 5 x ) d x$$

Evaluate

$$\frac{3}{4}=0.75$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}-6x^{2}+5x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -6x^{2}\mathrm{d}x+\int 5x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-6\int x^{2}\mathrm{d}x+5\int x\mathrm{d}x$$

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}.

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

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}.

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

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

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

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.

$$\frac{1^{4}}{4}-2\times 1^{3}+\frac{5}{2}\times 1^{2}-\left(\frac{0^{4}}{4}-2\times 0^{3}+\frac{5}{2}\times 0^{2}\right)$$

Simplify.

$$\frac{3}{4}$$