Solve ∫ (from - 5 to - 1) of 1/3x^4(6-x^5)^3 wrt x

$$\int _ { - 5 } ^ { - 1 } \frac { 1 } { 3 } x ^ { 4 } ( 6 - x ^ { 5 } ) ^ { 3 } d x$$

Evaluate

$$1601699426492$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(6-x^{5}\right)^{3}.

$$\int _{-5}^{-1}\frac{1}{3}x^{4}\left(216-108x^{5}+18\left(x^{5}\right)^{2}-\left(x^{5}\right)^{3}\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 5 and 2 to get 10.

$$\int _{-5}^{-1}\frac{1}{3}x^{4}\left(216-108x^{5}+18x^{10}-\left(x^{5}\right)^{3}\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 5 and 3 to get 15.

$$\int _{-5}^{-1}\frac{1}{3}x^{4}\left(216-108x^{5}+18x^{10}-x^{15}\right)\mathrm{d}x$$

Use the distributive property to multiply \frac{1}{3}x^{4} by 216-108x^{5}+18x^{10}-x^{15}.

$$\int _{-5}^{-1}72x^{4}-36x^{9}+6x^{14}-\frac{1}{3}x^{19}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 72x^{4}-36x^{9}+6x^{14}-\frac{x^{19}}{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 72x^{4}\mathrm{d}x+\int -36x^{9}\mathrm{d}x+\int 6x^{14}\mathrm{d}x+\int -\frac{x^{19}}{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$72\int x^{4}\mathrm{d}x-36\int x^{9}\mathrm{d}x+6\int x^{14}\mathrm{d}x-\frac{\int x^{19}\mathrm{d}x}{3}$$

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}. Multiply 72 times \frac{x^{5}}{5}.

$$\frac{72x^{5}}{5}-36\int x^{9}\mathrm{d}x+6\int x^{14}\mathrm{d}x-\frac{\int x^{19}\mathrm{d}x}{3}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{9}\mathrm{d}x with \frac{x^{10}}{10}. Multiply -36 times \frac{x^{10}}{10}.

$$\frac{72x^{5}}{5}-\frac{18x^{10}}{5}+6\int x^{14}\mathrm{d}x-\frac{\int x^{19}\mathrm{d}x}{3}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{14}\mathrm{d}x with \frac{x^{15}}{15}. Multiply 6 times \frac{x^{15}}{15}.

$$\frac{72x^{5}}{5}-\frac{18x^{10}}{5}+\frac{2x^{15}}{5}-\frac{\int x^{19}\mathrm{d}x}{3}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{19}\mathrm{d}x with \frac{x^{20}}{20}. Multiply -\frac{1}{3} times \frac{x^{20}}{20}.

$$\frac{72x^{5}}{5}-\frac{18x^{10}}{5}+\frac{2x^{15}}{5}-\frac{x^{20}}{60}$$

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.

$$\frac{72}{5}\left(-1\right)^{5}-\frac{18}{5}\left(-1\right)^{10}+\frac{2}{5}\left(-1\right)^{15}-\frac{\left(-1\right)^{20}}{60}-\left(\frac{72}{5}\left(-5\right)^{5}-\frac{18}{5}\left(-5\right)^{10}+\frac{2}{5}\left(-5\right)^{15}-\frac{\left(-5\right)^{20}}{60}\right)$$

Simplify.

$$1601699426492$$