Solve ∫ (from 4 to 9) of x^3+2x+1

$$\int _ { 4 } ^ { 9 } x ^ { 3 } + 2 x + 1$$

Evaluate

$$\frac{6585}{4}=1646.25$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}+2x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int 2x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+2\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{x^{4}}{4}+x^{2}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}+x^{2}+x$$

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.

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

Simplify.

$$\frac{6585}{4}$$