Solve ∫ (from 0 to 1) of x^2+7x+2

$$\int _ { 0 } ^ { 1 } x ^ { 2 } + 7 x + 2$$

Evaluate

$$\frac{35}{6}\approx 5.833333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+7x+2\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 7x\mathrm{d}x+\int 2\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+7\int x\mathrm{d}x+\int 2\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+7\int x\mathrm{d}x+\int 2\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+\frac{7x^{2}}{2}+\int 2\mathrm{d}x$$

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+\frac{7x^{2}}{2}+2x$$

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.

$$\frac{1^{3}}{3}+\frac{7}{2}\times 1^{2}+2\times 1-\left(\frac{0^{3}}{3}+\frac{7}{2}\times 0^{2}+2\times 0\right)$$

Simplify.

$$\frac{35}{6}$$