Solve ∫ (from 1 to 4) of (5-x+2x^2-3x^3) wrt x

$$\int _ { 1 } ^ { 4 } ( 5 - x + 2 x ^ { 2 } - 3 x ^ { 3 } ) d x$$

Evaluate

$$-\frac{567}{4}=-141.75$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}. Multiply -1 times \frac{x^{2}}{2}.

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

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}.

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

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

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

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.

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

Simplify.

$$-\frac{567}{4}$$