Solve ∫ (from 1 to 3) of (x^4+7x^3+2x^2+x+3) wrt x

$$\int _ { 1 } ^ { 3 } ( x ^ { 4 } + 7 x ^ { 3 } + 2 x ^ { 2 } + x + 3 ) d x$$

Evaluate

$$\frac{3236}{15}\approx 215.733333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{4}+7x^{3}+2x^{2}+x+3\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int 7x^{3}\mathrm{d}x+\int 2x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x+7\int x^{3}\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}+7\int x^{3}\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 7 times \frac{x^{4}}{4}.

$$\frac{x^{5}}{5}+\frac{7x^{4}}{4}+2\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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}. Multiply 2 times \frac{x^{3}}{3}.

$$\frac{x^{5}}{5}+\frac{7x^{4}}{4}+\frac{2x^{3}}{3}+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}+\frac{7x^{4}}{4}+\frac{2x^{3}}{3}+\frac{x^{2}}{2}+\int 3\mathrm{d}x$$

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}+\frac{7x^{4}}{4}+\frac{2x^{3}}{3}+\frac{x^{2}}{2}+3x$$

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.

$$\frac{3^{5}}{5}+\frac{7}{4}\times 3^{4}+\frac{2}{3}\times 3^{3}+\frac{3^{2}}{2}+3\times 3-\left(\frac{1^{5}}{5}+\frac{7}{4}\times 1^{4}+\frac{2}{3}\times 1^{3}+\frac{1^{2}}{2}+3\times 1\right)$$

Simplify.

$$\frac{3236}{15}$$