What is the generalized form of the integral $\int_0^2 x^3(1-x/2)^4dx$? I have tryed to solve it using the formula with euler, but i dont quite understand it.
Generalized form of the integral
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calculus
2 Answers
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Note
$$\int_0^2 x^3\left(1-\frac{x}{2}\right)^4\mathrm{d}x= \frac{1}{16}\int_{0}^{2} x^3(2-x)^4 \mathrm{d}x$$ Now, substitute $x=2t$. The question becomes $$\frac{2^8}{16} \times \int_{0}^{1} t^3(1-t)^4 \mathrm{d}t$$ Using the beta function, we can calcualte the result to be $$\frac{2^8}{2^4} \times \mathrm{B} (4,5)=\frac{3! \times 4!}{8!} \times 2^{4}=\frac{2}{35}$$
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Using S.C.B.'s first steps, you have $$\frac{2^8}{16} \times \int_{0}^{1} t^3(1-t)^4 \,dt=16\int_0 ^1(t^7-4 t^6+6 t^5-4 t^4+t^3)\,dt$$ which seems to be quite simple.