Basics

Calculus Level 3

∫ − 1 5 ( 1 + x ) ( 2 x + x 2 ) 5 d x = ? \displaystyle \int^{5}_{-1}(1+x)(2x+x^{2})^{5}dx =?

Note :You can use calculator at final step of your calculation.


The answer is 153188802.

Akshat Sharda by Akshat Sharda , 21, India

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3 solutions 

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let 
u = x^2 + 2x

du = 2(x+1) dx

Then the integral becomes:
1/2 integral of u^5 from u=-1 to u=35

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution.What most people forget to do is that when you are taking substitution , since it is definite integration the limits will also change according to the substitution.Upvoted your solution

Athiyaman Nallathambi - 5 years, 9 months ago

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Yeah....you are correct :-)

Akshat Sharda - 5 years, 9 months ago

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Ive comitted so many countless mistakes and this is one of them.Thats why I take caution before solving any question.

Athiyaman Nallathambi - 5 years, 9 months ago

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@Athiyaman Nallathambi – Yeah.....I also do a lot of mistakes and very few times I really learn from them ..... very few ^_^

Akshat Sharda - 5 years, 9 months ago

to avoid those mistakes i always substitute u back to original

that is u to 2x+x^2

Akash singh - 5 years, 9 months ago

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Even I do the same.

Akshat Sharda - 5 years, 9 months ago

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@Akshat Sharda – great.........^_^

Akash singh - 5 years, 9 months ago

Look at my solution now.

Akshat Sharda - 5 years, 9 months ago
Akshat Sharda
Aug 20, 2015

∫ − 1 5 ( 1 + x ) ( 2 x + x 2 ) 5 d x \displaystyle \int^{5}_{-1} (1+x)\color{#3D99F6}{(2x+x^{2})}^{5}dx

Let ( 2 x + x 2 ) = u \color{#3D99F6}{(2x+x^{2})} =\color{#D61F06}{u} ,

⇒ ∫ − 1 5 ( 1 + x ) ( 2 x + x 2 ) 5 d x \Rightarrow \displaystyle \int^{5}_{-1} (1+x)\color{#3D99F6}{(2x+x^{2})}^{5}dx

⇒ ∫ − 1 5 ( 1 + x ) u 5 d x \Rightarrow \displaystyle \int^{5}_{-1} (1+x)\color{#D61F06}{u}^{5}dx

Now we can re-write it as ,

⇒ ∫ − 1 5 1 2 ( 1 + x ) ( 1 + x ) u 5 d x × d u d x \Rightarrow \displaystyle \int^{5}_{-1} \color{#D61F06}{\frac{1}{2(1+x)}}(1+x)\color{#D61F06}{u}^{5}dx×\color{#D61F06}{\frac{du}{dx}}

⇒ ∫ − 1 5 1 2 u 5 d u \Rightarrow \displaystyle \int^{5}_{-1}\frac12u^{5}\color{#D61F06}{du}

⇒ 1 2 5 + 1 u 5 + 1 \Rightarrow \displaystyle \frac{\frac{1}{2}}{5+1}\color{#D61F06}{u}^{5+1} ] − 1 5 ]_{-1}^{5}

⇒ 1 12 u 6 \Rightarrow \displaystyle \frac{1}{12}\color{#D61F06}{u}^{6} ] − 1 5 ]^{5}_{-1}

Replacing u \color{#D61F06}{u} with ( 2 x + x 2 ) \color{#3D99F6}{(2x+x^{2})} ,

⇒ 1 12 ( 2 x + x 2 ) 6 \Rightarrow \displaystyle \frac{1}{12}\color{#3D99F6}{(2x+x^{2})}^{6} ] − 1 5 ]^{5}_{-1}

⇒ 3 5 6 12 − − 1 6 12 \Rightarrow \frac{35^{6}}{12} - \frac{-1^{6}}{12}

⇒ 153188802 \Rightarrow \huge \color{#EC7300}{\boxed{153188802}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Akash Singh
Aug 20, 2015

take 2x+x^2=t so it becomes as

integration[(t^5dt)/2]

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