Solve ∫ (from 0 to 1) of (2x+1) wrt x

$$\int _ { 0 } ^ { 1 } ( 2 x + 1 ) d x$$

Evaluate

$$2$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

$$x^{2}+x$$

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.

$$1^{2}+1-\left(0^{2}+0\right)$$

Simplify.

$$2$$