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\int _{0}^{\frac{1}{2}}x^{4}\left(1-6x+12x^{2}-8x^{3}\right)\mathrm{d}x
Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(1-2x\right)^{3}.
\int _{0}^{\frac{1}{2}}x^{4}-6x^{5}+12x^{6}-8x^{7}\mathrm{d}x
Use the distributive property to multiply x^{4} by 1-6x+12x^{2}-8x^{3}.
\int x^{4}-6x^{5}+12x^{6}-8x^{7}\mathrm{d}x
Evaluate the indefinite integral first.
\int x^{4}\mathrm{d}x+\int -6x^{5}\mathrm{d}x+\int 12x^{6}\mathrm{d}x+\int -8x^{7}\mathrm{d}x
Integrate the sum term by term.
\int x^{4}\mathrm{d}x-6\int x^{5}\mathrm{d}x+12\int x^{6}\mathrm{d}x-8\int x^{7}\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{5}}{5}-6\int x^{5}\mathrm{d}x+12\int x^{6}\mathrm{d}x-8\int x^{7}\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}-x^{6}+12\int x^{6}\mathrm{d}x-8\int x^{7}\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply -6 times \frac{x^{6}}{6}.
\frac{x^{5}}{5}-x^{6}+\frac{12x^{7}}{7}-8\int x^{7}\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply 12 times \frac{x^{7}}{7}.
\frac{x^{5}}{5}-x^{6}+\frac{12x^{7}}{7}-x^{8}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{7}\mathrm{d}x with \frac{x^{8}}{8}. Multiply -8 times \frac{x^{8}}{8}.
-x^{8}+\frac{12x^{7}}{7}-x^{6}+\frac{x^{5}}{5}
Simplify.
-\left(\frac{1}{2}\right)^{8}+\frac{12}{7}\times \left(\frac{1}{2}\right)^{7}-\left(\frac{1}{2}\right)^{6}+\frac{\left(\frac{1}{2}\right)^{5}}{5}-\left(-0^{8}+\frac{12}{7}\times 0^{7}-0^{6}+\frac{0^{5}}{5}\right)
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{1}{8960}
Simplify.