Integration of 3x (x-1)³
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Answer:
I=∫x(1−x)3dx ; Substitute 1−x=t ; dx=−dt
=−∫(1−t)t3dt
=−∫(t3−t4)dt
=−(t44−t55)+C
=(1−x)55−(1−x)44+C
The method used above is too long.Simpler is,
I=∫x(1−x)3dx
=∫{1−(1−x)}(1−x)3dx
=∫{(1−x)3−(1−x)4}dx
=−(1−x)44+(1−x)55+C
=(1−x)55−(1−x)44+C
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Answer:
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