Question

Find the integration of the function given $\int{\left( 1-x \right)\sqrt{x}dx}$.

Answer 1

We start solving the problem by substituting ${{y}^{2}}$ in place of x in the integral $\int{\left( 1-x \right)\sqrt{x}dx}$. After substituting we make use of the formulas of integration $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$in order to get the answer in variables of y. We then convert the obtained answer in terms of x from y to get the required result.

Complete step by step answer:
According to the problem, it is given that we need to find the integration of a given function $\int{\left( 1-x \right)\sqrt{x}dx}$.
We have got $\int{\left( 1-x \right)\sqrt{x}dx}$ ---(1).
Let us assume $x={{y}^{2}}$ ---(2).
On differentiating both sides, we get $dx=d\left( {{y}^{2}} \right)$.
$dx=2ydy$---(3).
So, we substitute equations (2) and (3) in equation (1).
So, we have got $\int{\left( 1-x \right)\sqrt{x}dx}=\int{\left( 1-{{y}^{2}} \right).\sqrt{{{y}^{2}}}.2ydy}$.
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\int{\left( 1-{{y}^{2}} \right).y.2ydy}$ ---(4).
We know that $\int{af\left( x \right)dx=a\int{f\left( x \right)dx}}$. We use this result in equation (4).
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=2.\int{\left( 1-{{y}^{2}} \right).{{y}^{2}}dy}$.
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=2.\int{\left( {{y}^{2}}-{{y}^{4}} \right)dy}$ ---(5).
We know that $\int{\left( a\left( x \right)+b\left( x \right) \right)dx=\int{a\left( x \right)dx+\int{b\left( x \right)dx}}}$. We use this result in equation (5)
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=2\left( \int{{{y}^{2}}dy}-\int{{{y}^{4}}dy} \right)$ ---(6).
We know that the integration is defined as$\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$. We use this result in equation (6).
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=2\left( \dfrac{{{y}^{3}}}{3}-\dfrac{{{y}^{5}}}{5} \right)+C$.
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\dfrac{2{{y}^{3}}}{3}-\dfrac{2{{y}^{5}}}{5}+C$ ---(7).
From equation (2), We have got $x={{y}^{2}}$.
We have got $y=\sqrt{x}$.
We have got $y={{x}^{\dfrac{1}{2}}}$ ---(8).
We substitute the result of equation (8) in equation (7).
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\dfrac{2{{\left( {{x}^{\dfrac{1}{2}}} \right)}^{3}}}{3}-\dfrac{2{{\left( {{x}^{\dfrac{1}{2}}} \right)}^{5}}}{5}+C$ ---(9).
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$, we use this result in equation (9).
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\dfrac{2{{x}^{\dfrac{3}{2}}}}{3}-\dfrac{2{{x}^{\dfrac{5}{2}}}}{5}+C$.
We have found the integration of the function $\int{\left( 1-x \right)\sqrt{x}dx}$ as $\dfrac{2{{x}^{\dfrac{3}{2}}}}{3}-\dfrac{2{{x}^{\dfrac{5}{2}}}}{5}+C$.
∴ The integration of the function $\int{\left( 1-x \right)\sqrt{x}dx}$ is $\dfrac{2{{x}^{\dfrac{3}{2}}}}{3}-\dfrac{2{{x}^{\dfrac{5}{2}}}}{5}+C$.

Note:
We should not forget to add the arbitrary constant after integrating any function. Alternatively, we can solve the problem as follows:
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\int{\left( {{x}^{\dfrac{1}{2}}}-{{x}^{\dfrac{3}{2}}} \right)}dx$.
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\int{{{x}^{\dfrac{1}{2}}}dx-\int{{{x}^{\dfrac{3}{2}}}}dx}$.
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}-\dfrac{{{x}^{\dfrac{3}{2}+1}}}{\dfrac{3}{2}+1}+C$.
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\dfrac{{{x}^{\dfrac{5}{2}}}}{\dfrac{5}{2}}+C$.
We have got $\int{\left( 1-x \right)\sqrt{x}dx}=\dfrac{2{{x}^{\dfrac{3}{2}}}}{3}-\dfrac{2{{x}^{\dfrac{5}{2}}}}{5}+C$.
∴ The integration of the function $\int{\left( 1-x \right)\sqrt{x}dx}$ is $\dfrac{2{{x}^{\dfrac{3}{2}}}}{3}-\dfrac{2{{x}^{\dfrac{5}{2}}}}{5}+C$.