Solve ∫ (from 1 to 4) of x^2+1 wrt x

$$\int_{ 1 }^{ 4 } { x }^{ 2 } +1 d x$$

Evaluate

$$24$$

Solution Steps

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.

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

Simplify.

$$24$$