Solve ∫ (from 1 to 2) of x^2-9x+61/6 wrt x

$$\int _ { 1 } ^ { 2 } x ^ { 2 } - 9 x + \frac { 61 } { 6 } d x$$

Evaluate

$$-1$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}-9x+\frac{61}{6}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -9x\mathrm{d}x+\int \frac{61}{6}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-9\int x\mathrm{d}x+\int \frac{61}{6}\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-9\int x\mathrm{d}x+\int \frac{61}{6}\mathrm{d}x$$

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 -9 times \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}-\frac{9x^{2}}{2}+\int \frac{61}{6}\mathrm{d}x$$

Find the integral of \frac{61}{6} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-\frac{9x^{2}}{2}+\frac{61x}{6}$$

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.

$$\frac{2^{3}}{3}-\frac{9}{2}\times 2^{2}+\frac{61}{6}\times 2-\left(\frac{1^{3}}{3}-\frac{9}{2}\times 1^{2}+\frac{61}{6}\times 1\right)$$

Simplify.

$$-1$$