Solve ∫ (from 0 to 5) of x^2(x-3)(x-5)

$$\int _ { 0 } ^ { 5 } x ^ { 2 } ( x - 3 ) ( x - 5 )$$

Evaluate

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Solution Steps

Use the distributive property to multiply x^{2} by x-3.

$$\int _{0}^{5}\left(x^{3}-3x^{2}\right)\left(x-5\right)\mathrm{d}x$$

Use the distributive property to multiply x^{3}-3x^{2} by x-5 and combine like terms.

$$\int _{0}^{5}x^{4}-8x^{3}+15x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{4}-8x^{3}+15x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -8x^{3}\mathrm{d}x+\int 15x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-8\int x^{3}\mathrm{d}x+15\int x^{2}\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-8\int x^{3}\mathrm{d}x+15\int x^{2}\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-2x^{4}+15\int x^{2}\mathrm{d}x$$

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

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

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.

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

Simplify.

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