Solve ∫ (from 0 to frac { 1) of {3}}(2x^2+4x-10) wrt x

$$\int _ { 0 } ^ { \frac { 1 } { 3 } } ( 2 x ^ { 2 } + 4 x - 10 ) d x$$

Evaluate

$$-\frac{250}{81}\approx -3.086419753$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x^{2}+4x-10\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int 4x\mathrm{d}x+\int -10\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x+4\int x\mathrm{d}x+\int -10\mathrm{d}x$$

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 2 times \frac{x^{3}}{3}.

$$\frac{2x^{3}}{3}+4\int x\mathrm{d}x+\int -10\mathrm{d}x$$

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}.

$$\frac{2x^{3}}{3}+2x^{2}+\int -10\mathrm{d}x$$

Find the integral of -10 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{3}+2x^{2}-10x$$

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.

$$\frac{2}{3}\times \left(\frac{1}{3}\right)^{3}+2\times \left(\frac{1}{3}\right)^{2}-10\times \frac{1}{3}-\left(\frac{2}{3}\times 0^{3}+2\times 0^{2}-10\times 0\right)$$

Simplify.

$$-\frac{250}{81}$$