Solve ∫ (from 0 to 20) of {left(11-9/20xright)}^210 wrt x

$$\int_{ 0 }^{ 20 } { \left( 11- \frac{ 9 }{ 20 } x \right) }^{ 2 } 10 d x$$

Evaluate

$$9800$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(11-\frac{9}{20}x\right)^{2}.

$$\int _{0}^{20}\left(121-\frac{99}{10}x+\frac{81}{400}x^{2}\right)\times 10\mathrm{d}x$$

Use the distributive property to multiply 121-\frac{99}{10}x+\frac{81}{400}x^{2} by 10.

$$\int _{0}^{20}1210-99x+\frac{81}{40}x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 1210-99x+\frac{81x^{2}}{40}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1210\mathrm{d}x+\int -99x\mathrm{d}x+\int \frac{81x^{2}}{40}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 1210\mathrm{d}x-99\int x\mathrm{d}x+\frac{81\int x^{2}\mathrm{d}x}{40}$$

Find the integral of 1210 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$1210x-99\int x\mathrm{d}x+\frac{81\int x^{2}\mathrm{d}x}{40}$$

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 -99 times \frac{x^{2}}{2}.

$$1210x-\frac{99x^{2}}{2}+\frac{81\int x^{2}\mathrm{d}x}{40}$$

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 \frac{81}{40} times \frac{x^{3}}{3}.

$$1210x-\frac{99x^{2}}{2}+\frac{27x^{3}}{40}$$

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.

$$1210\times 20-\frac{99}{2}\times 20^{2}+\frac{27}{40}\times 20^{3}-\left(1210\times 0-\frac{99}{2}\times 0^{2}+\frac{27}{40}\times 0^{3}\right)$$

Simplify.

$$9800$$