integration of (π-x)^2
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check the attachment
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jadensingh:
answer is (π-3)^3÷(-3)
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∫(π−x)2 dx=−∫(π−x)2 (−dx)
=−∫(π−x)2 d(π−x)
=−(π−x)33+C
Alternatively,
∫(π−x)2 dx=∫(π2+x2−2πx) dx
=π2x+x33−2πx22+C
=π2x+x33−πx2+C
I don't know hindi
speak in English
ok how you take out minus in starting
it is common right
why u take common
my problem is that when u intergrate power which is 2 then its 2+1=3upon 3 means (π-3)3÷3
but answer is (π-3)^3÷-3
how -3
sorry actually its π-x
in place of 3
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