Solve ∫ (from 0 to 2) of 2/3x^2-5/4x^3 wrt x

$$\int _ { 0 } ^ { 2 } \frac { 2 } { 3 } x ^ { 2 } - \frac { 5 } { 4 } x ^ { 3 } d x$$

Evaluate

$$-\frac{29}{9}\approx -3.222222222$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$\frac{2x^{3}}{9}-\frac{5\int x^{3}\mathrm{d}x}{4}$$

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 -\frac{5}{4} times \frac{x^{4}}{4}.

$$\frac{2x^{3}}{9}-\frac{5x^{4}}{16}$$

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.

$$\frac{2}{9}\times 2^{3}-\frac{5}{16}\times 2^{4}-\left(\frac{2}{9}\times 0^{3}-\frac{5}{16}\times 0^{4}\right)$$

Simplify.

$$-\frac{29}{9}$$