Evaluate the value of the given integral $\int {\left( {1 + x} \right)\sqrt x dx} $.
Answer
362.4k+ views
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.
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
•
Total views: 362.4k
•
Views today: 4.62k
Recently Updated Pages
What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

The poet says Beauty is heard in Can you hear beauty class 6 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the past tense of read class 10 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
