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

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

Evaluate

$$\frac{10}{3}\approx 3.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{10}{3}$$