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

$$\int _ { 0 } ^ { 1 } x ^ { 2 } - 3 x ^ { 3 } + 2 x ^ { 4 } d x$$

Evaluate

$$-\frac{1}{60}\approx -0.016666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 -3 times \frac{x^{4}}{4}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 2 times \frac{x^{5}}{5}.

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

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.

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

Simplify.

$$-\frac{1}{60}$$