Solve ∫ (from pi to x _ { 2) of }x+1

$$\int _ { \pi } ^ { x _ { 2 } } x + 1$$

Evaluate

$$\frac{x_{2}^{2}+2x_{2}-\pi \left(\pi +2\right)}{2}$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

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

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

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.

$$\frac{x_{2}^{2}}{2}+x_{2}-\left(\frac{\pi ^{2}}{2}+\pi \right)$$

Simplify.

$$\frac{x_{2}^{2}}{2}+x_{2}-\frac{\pi \left(\pi +2\right)}{2}$$

Differentiate w.r.t. x_2

$$x_{2}+1$$