Solve ∫ (from 0 to 3) of (x^4-8x^3+5x-10) wrt x

$$\int _ { 0 } ^ { 3 } ( x ^ { 4 } - 8 x ^ { 3 } + 5 x - 10 ) d x$$

Evaluate

$$-\frac{1209}{10}=-120.9$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{4}-8x^{3}+5x-10\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -8x^{3}\mathrm{d}x+\int 5x\mathrm{d}x+\int -10\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-8\int x^{3}\mathrm{d}x+5\int x\mathrm{d}x+\int -10\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-8\int x^{3}\mathrm{d}x+5\int x\mathrm{d}x+\int -10\mathrm{d}x$$

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 -8 times \frac{x^{4}}{4}.

$$\frac{x^{5}}{5}-2x^{4}+5\int x\mathrm{d}x+\int -10\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-2x^{4}+\frac{5x^{2}}{2}+\int -10\mathrm{d}x$$

Find the integral of -10 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}-2x^{4}+\frac{5x^{2}}{2}-10x$$

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.

$$\frac{3^{5}}{5}-2\times 3^{4}+\frac{5}{2}\times 3^{2}-10\times 3-\left(\frac{0^{5}}{5}-2\times 0^{4}+\frac{5}{2}\times 0^{2}-10\times 0\right)$$

Simplify.

$$-\frac{1209}{10}$$