Solve ∫ (from 1 to 2) of (1/3x^3-3/2x^2) wrt x

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

Evaluate

$$-\frac{9}{4}=-2.25$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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{1}{3} times \frac{x^{4}}{4}.

$$\frac{x^{4}}{12}-\frac{3\int x^{2}\mathrm{d}x}{2}$$

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

$$\frac{x^{4}}{12}-\frac{x^{3}}{2}$$

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.

$$\frac{2^{4}}{12}-\frac{2^{3}}{2}-\left(\frac{1^{4}}{12}-\frac{1^{3}}{2}\right)$$

Simplify.

$$-\frac{9}{4}$$