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integral limits from 0 to 1 x^2(1+x)^2dx= integral limits from 0 to 1 x^2(1+x)^2dx=
0 to 1 integral x^2(1+x)^2dx =0 to 1 integral x^2(1+x^2+2x)dx =0to 1 integral (x^2+x^4+2x^3)dx =0 to 1 integral x^2dx + 0 to 1 integral x^4dx + 2* 0 to 1 integral x^3dx =0 to 1 [x^3/3] + 0 to 1 [x^5/5] + 2* 0 to 1 [x^4/4] =[1/3-0/3] + [1/5-0/5] + 2*[1/4-0/4] =1/3 + 1/5 + 2*1/4 =1/3 + 1/5 + 1/2= 31/30
0 to 1 integral x^2(1+x)^2dx
=0 to 1 integral x^2(1+x^2+2x)dx
=0to 1 integral (x^2+x^4+2x^3)dx
=0 to 1 integral x^2dx + 0 to 1 integral x^4dx + 2* 0 to 1 integral x^3dx
=0 to 1 [x^3/3] + 0 to 1 [x^5/5] + 2* 0 to 1 [x^4/4]
=[1/3-0/3] + [1/5-0/5] + 2*[1/4-0/4]
=1/3 + 1/5 + 2*1/4
=1/3 + 1/5 + 1/2= 31/30
are yar u just expand (x+1)2 and multiply by x2 u will get x2 +x4 +2x3 now on integrating u will get x3/3 + x5/5 + x4/2 and on putting limits u will get 1/3+1/5+1/2=31/30 good luck
are yar
u just expand (x+1)2
and multiply by x2
u will get x2 +x4 +2x3
now on integrating u will get x3/3 + x5/5 + x4/2 and on putting limits
u will get 1/3+1/5+1/2=31/30
good luck
i think the ans is 31/30....."is it right??"
Yes 31/30 is right.
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