Solve ∫ (from 1 to 10) of -20x^3+240x^2-1200 wrt x

$$\int _ { 1 } ^ { 10 } - 20 x ^ { 3 } + 240 x ^ { 2 } - 1200 d x$$

Evaluate

$$19125$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -20x^{3}+240x^{2}-1200\mathrm{d}x$$

Integrate the sum term by term.

$$\int -20x^{3}\mathrm{d}x+\int 240x^{2}\mathrm{d}x+\int -1200\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-20\int x^{3}\mathrm{d}x+240\int x^{2}\mathrm{d}x+\int -1200\mathrm{d}x$$

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 -20 times \frac{x^{4}}{4}.

$$-5x^{4}+240\int x^{2}\mathrm{d}x+\int -1200\mathrm{d}x$$

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 240 times \frac{x^{3}}{3}.

$$-5x^{4}+80x^{3}+\int -1200\mathrm{d}x$$

Find the integral of -1200 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-5x^{4}+80x^{3}-1200x$$

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.

$$-5\times 10^{4}+80\times 10^{3}-1200\times 10-\left(-5\times 1^{4}+80\times 1^{3}-1200\right)$$

Simplify.

$$19125$$