Solve ∫ (from 3 to 4) of -x^2+7x-12 wrt x

$$\int_{ 3 }^{ 4 } - { x }^{ 2 } +7x-12 d x$$

Evaluate

$$\frac{1}{6}\approx 0.166666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -x^{2}+7x-12\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{2}\mathrm{d}x+\int 7x\mathrm{d}x+\int -12\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{2}\mathrm{d}x+7\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$-\frac{x^{3}}{3}+7\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$-\frac{x^{3}}{3}+\frac{7x^{2}}{2}+\int -12\mathrm{d}x$$

Find the integral of -12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x^{3}}{3}+\frac{7x^{2}}{2}-12x$$

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.

$$-\frac{4^{3}}{3}+\frac{7}{2}\times 4^{2}-12\times 4-\left(-\frac{3^{3}}{3}+\frac{7}{2}\times 3^{2}-12\times 3\right)$$

Simplify.

$$\frac{1}{6}$$