Solve ∫ (from -4 to -1) of (x^2+9x+10) wrt x

$$\int_{ -4 }^{ -1 } ( { x }^{ 2 } +9x+10) d x$$

Evaluate

$$-\frac{33}{2}=-16.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+9x+10\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 9x\mathrm{d}x+\int 10\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+9\int x\mathrm{d}x+\int 10\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+9\int x\mathrm{d}x+\int 10\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+\frac{9x^{2}}{2}+\int 10\mathrm{d}x$$

Find the integral of 10 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+\frac{9x^{2}}{2}+10x$$

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.

$$\frac{\left(-1\right)^{3}}{3}+\frac{9}{2}\left(-1\right)^{2}+10\left(-1\right)-\left(\frac{\left(-4\right)^{3}}{3}+\frac{9}{2}\left(-4\right)^{2}+10\left(-4\right)\right)$$

Simplify.

$$-\frac{33}{2}$$