Solve ∫ (from 0 to 1) of (x^2+5x+1) wrt x

$$\int _ { 0 } ^ { 1 } ( x ^ { 2 } + 5 x + 1 ) d x$$

Evaluate

$$\frac{23}{6}\approx 3.833333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+5x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 5x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+5\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+5\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+\frac{5x^{2}}{2}+\int 1\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+\frac{5x^{2}}{2}+x$$

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.

$$\frac{1^{3}}{3}+\frac{5}{2}\times 1^{2}+1-\left(\frac{0^{3}}{3}+\frac{5}{2}\times 0^{2}+0\right)$$

Simplify.

$$\frac{23}{6}$$