Solve ∫ (from 0 to 3) of 2/7x^4-3x^3+2x+5 wrt x

$$\int _ { 0 } ^ { 3 } \frac { 2 } { 7 } x ^ { 4 } - 3 x ^ { 3 } + 2 x + 5 d x$$

Evaluate

$$-\frac{3201}{140}\approx -22.864285714$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}. Multiply \frac{2}{7} times \frac{x^{5}}{5}.

$$\frac{2x^{5}}{35}-3\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x+\int 5\mathrm{d}x$$

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 -3 times \frac{x^{4}}{4}.

$$\frac{2x^{5}}{35}-\frac{3x^{4}}{4}+2\int x\mathrm{d}x+\int 5\mathrm{d}x$$

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}.

$$\frac{2x^{5}}{35}-\frac{3x^{4}}{4}+x^{2}+\int 5\mathrm{d}x$$

Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{5}}{35}-\frac{3x^{4}}{4}+x^{2}+5x$$

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.

$$\frac{2}{35}\times 3^{5}-\frac{3}{4}\times 3^{4}+3^{2}+5\times 3-\left(\frac{2}{35}\times 0^{5}-\frac{3}{4}\times 0^{4}+0^{2}+5\times 0\right)$$

Simplify.

$$-\frac{3201}{140}$$