Solve ∫ (from 1 to 4) of (x^2/2-3x+9/2) wrt x

$$\int _ { 1 } ^ { 4 } ( \frac { x ^ { 2 } } { 2 } - 3 x + \frac { 9 } { 2 } ) d x$$

Evaluate

$$\frac{3}{2}=1.5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 \frac{1}{2} times \frac{x^{3}}{3}.

$$\frac{x^{3}}{6}-3\int x\mathrm{d}x+\int \frac{9}{2}\mathrm{d}x$$

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

$$\frac{x^{3}}{6}-\frac{3x^{2}}{2}+\int \frac{9}{2}\mathrm{d}x$$

Find the integral of \frac{9}{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{6}-\frac{3x^{2}}{2}+\frac{9x}{2}$$

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.

$$\frac{4^{3}}{6}-\frac{3}{2}\times 4^{2}+\frac{9}{2}\times 4-\left(\frac{1^{3}}{6}-\frac{3}{2}\times 1^{2}+\frac{9}{2}\times 1\right)$$

Simplify.

$$\frac{3}{2}$$