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

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

Evaluate

$$\frac{62}{3}\approx 20.666666667$$

Solution Steps

Subtract 2 from 4 to get 2.

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

Multiply 2 and \frac{1}{2} to get 1.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

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

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

$$\frac{x^{3}}{3}+x$$

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.

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

Simplify.

$$\frac{62}{3}$$