Question

Find the following integrals ∫(1−x)x​dx.

Answer 1

Solution:

Given Integral:

$\displaystyle\tt\longrightarrow I=\int(1-x)x\: dx$

On multiplying, we get:

$\displaystyle\tt\longrightarrow I=\int(x-x^{2})\: dx$

Using properties of integrals, we can write:

$\displaystyle\tt\longrightarrow I=\int x\:dx-\int x^{2}\: dx$

As we know that:

$\bigstar\:\:\underline{\boxed{\tt\int x^{n}\:dx=\dfrac{x^{n+1}}{n+1}+C}}$

Using this result, we get:

$\displaystyle\tt\longrightarrow I=\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}+C$

Therefore:

$\displaystyle\tt\longrightarrow \int (1-x)x\:dx=\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}+C$

Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$