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Evaluate the value of the given integral $\int {\left( {1 + x} \right)\sqrt x dx} $.

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Hint: In this question first we will multiply the given two functions then apply power rule of Integration which says that integration of\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\] where $n \ne - 1$. As if $n = - 1$then ${x^{ - 1}} = \dfrac{1}{x}$ and Integration of $\int {\dfrac{1}{x}dx = \ln x + c} $, where c is the integration constant.

Complete step-by-step answer:
In the given question the integral is an indefinite type of integration as there are no upper and lower limits. To solve this question first of all, expand the given expression by multiplying $\sqrt x $.
\[ \Rightarrow \left( {1 + x} \right)\sqrt x = \left( {\sqrt x + {x^1} \times {x^{\dfrac{1}{2}}}} \right)\]
The above expression can be simplified more such that
   \Rightarrow \left( {1 + x} \right)\sqrt x = \left( {\sqrt x + {x^{\dfrac{1}{2} + 1}}} \right) \\
   \Rightarrow \left( {1 + x} \right)\sqrt x = \left( {\sqrt x + {x^{\dfrac{3}{2}}}} \right) \\
Now the Integration can be done easily
$ \Rightarrow \int {\left( {\sqrt x + {x^{\dfrac{3}{2}}}} \right)dx} $
Integrating each term separately as integral of a sum is the sum of the integrals.
$ \Rightarrow \int {\sqrt x dx + \int {{x^{\dfrac{3}{2}}}} dx} $
Now using the Power rule of integration \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \],we get
\[ \Rightarrow \dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} + \dfrac{{{x^{\dfrac{5}{2}}}}}{{\dfrac{5}{2}}} + c\]
Further simplify the above expression
\[ \Rightarrow \dfrac{{2{x^{\dfrac{3}{2}}}}}{3} + \dfrac{{2{x^{\dfrac{5}{2}}}}}{5} + c\]

Note: Whenever this type of question appears always first write down the given expression which needs to be integrated. Afterwards try to simplify the expression by expanding as much as possible (as in our case we simplified \[\left( {1 + x} \right)\sqrt x \]).Use power rule of integration which says that integration of\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\], Where c is the integration constant need to be kept at the end of integration as differentiation of Constant is zero. Differentiation actually eats the constant values because of this differentiation and integration are not exactly inverse operation of each other. Do understand the basic rules of integration as it's going to help further solving the tough question with more tough expressions.
Last updated date: 25th Sep 2023
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