Question

Evaluate the following: $\int{\left( 1-x \right)}\sqrt{x}dx$.
(a) $\dfrac{2}{3}{{x}^{{}^{3}/{}_{2}}}-\dfrac{2}{5}{{x}^{{}^{5}/{}_{2}}}+c$
(b) $\dfrac{2}{3}{{x}^{{}^{5}/{}_{2}}}-\dfrac{2}{5}{{x}^{{}^{3}/{}_{2}}}+c$
(c) $\dfrac{2}{3}{{x}^{{}^{3}/{}_{2}}}+\dfrac{2}{5}{{x}^{{}^{5}/{}_{2}}}+c$
(d) $\dfrac{2}{3}{{x}^{{}^{5}/{}_{2}}}+\dfrac{2}{5}{{x}^{{}^{3}/{}_{2}}}+c$

Answer 1

Hint: In order to solve these types of integration, the best way to is to reduce the given form into standard form. The standard form is given by $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$. On reducing the given integration in this form, we can easily evaluate the given integration. Besides, if the question has some upper limit and lower limit. The best way to solve the same integration with upper limit and lower limit is to remove the upper limit and lower limit and reduce the integration into standard form as shown above. After changing the given integration into standard form, we need to evaluate the equation using the formula given above. After evaluating the integration of standard form, now we can easily apply upper and lower limits and then solve the expression after applying the limits and hence, in this way we can solve the same question with upper limit and lower limit. In this question, we will convert the given integration into standard form by using the numerical value of the square root we will solve it.

Complete step by step answer:
Here, we are given the integration as $I=\int{\left( 1-x \right)}\sqrt{x}dx\text{ (say) }..............\text{(i)}$. We have to evaluate the given integration equation (i). In order to evaluate the given integration in equation (i), we will have to express in the standard form of $\int{{{x}^{n}}dx}$.
So, on expressing the integration given in equation (i) into standard form, we get,
$\Rightarrow I=\int{\left( 1-x \right)}\sqrt{x}dx$
$\Rightarrow I=\int{\left( 1-x \right)}{{\left( x \right)}^{{}^{1}/{}_{2}}}dx$
$\Rightarrow I=\int{{{\left( x \right)}^{{}^{1}/{}_{2}}}\left( 1-x \right)}dx$
\[\Rightarrow I=\int{\left( {{\left( x \right)}^{{}^{1}/{}_{2}}}-x.{{\left( x \right)}^{{}^{1}/{}_{2}}} \right)}dx\]
We can use index property which says we can add the power if the base is same, so, we get,
\[\Rightarrow I=\int{\left( {{\left( x \right)}^{{}^{1}/{}_{2}}}-{{\left( x \right)}^{1+\left( {}^{1}/{}_{2} \right)}} \right)}dx\]
\[\Rightarrow I=\int{\left( {{\left( x \right)}^{{}^{1}/{}_{2}}}-{{\left( x \right)}^{{}^{3}/{}_{2}}} \right)}dx\]
\[\Rightarrow I=\int{{{\left( x \right)}^{{}^{1}/{}_{2}}}}dx-\int{{{\left( x \right)}^{{}^{3}/{}_{2}}}}dx............(ii)\]
Now, the equation obtained in equation (ii) is in the standard form. Now, we can easily use the formula of standard form in order to evaluate the integration in equation (ii). The formula for standard form is given by –
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$ where c is the integration constant.
Hence, we use the standard form formula to evaluate the equation in equation (ii), we get,
\[\Rightarrow I=\dfrac{{{\left( x \right)}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}-\dfrac{{{\left( x \right)}^{\dfrac{3}{2}+1}}}{\dfrac{3}{2}+1}+c\]
\[\Rightarrow I=\dfrac{{{\left( x \right)}^{\dfrac{1+2}{2}}}}{\dfrac{1+2}{2}}-\dfrac{{{\left( x \right)}^{\dfrac{3+2}{2}}}}{\dfrac{3+2}{2}}+c\]
\[\Rightarrow I=\dfrac{{{\left( x \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\dfrac{{{\left( x \right)}^{\dfrac{5}{2}}}}{\dfrac{5}{2}}+c\]
\[\therefore I=\dfrac{2}{3}{{\left( x \right)}^{\dfrac{3}{2}}}-\dfrac{2}{5}{{\left( x \right)}^{\dfrac{5}{2}}}+c\]
Thus, the required answer is \[\dfrac{2}{3}{{\left( x \right)}^{\dfrac{3}{2}}}-\dfrac{2}{5}{{\left( x \right)}^{\dfrac{5}{2}}}+c\] for the given integration in equation (i).

So, the correct answer is “Option A”.

Note: For the above types of question, it is very important that a student remembers the standard from and the respective formula. As the above given integration can easily be solved by changing the given equation into the standard form of $\int{{{x}^{n}}dx}$ and then applying the formula of standard form as $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$. Besides, students often make mistakes while the standard form formula and evaluating the expression. Students should be careful while representing the standard form.