11Ans 2) - 1/2 ????
1)\int_{0}^{2a}\sqrt{2ax-x^{2}}
2)\int_{-1/2}^{1/2}\left ( [x]+ln(\frac{1+x}{1-x}) \right )dx
3)\int_{-2}^{2}\frac{3x^{5}+4x^{3}+2x^{2}+x+20}{x^{2}+4}
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6 Answers
211) given integral is ∫ √a2-(x-a)2 ....take x-a =t
its a standard form of integral.....hope u can do it after that
11Ans 2) Since ln (1+x1-x) is an odd function,
Therefore, -1/2 ∫ 1/2 { [x] + ln (1+x1-x)) dx
= -1/2 ∫ 0 [x] dx + 0 ∫ 1/2 [x] dx
= -1 (0+1/2) + 0
= -1/2
111)
\int_{0}^{2a}\sqrt{2ax-x^{2}}
u can write it as
2\int_{0}^{a}{\sqrt{x}\sqrt{2a-x}}dx
put a-x =t
den u get x=a-t
so
u have
2\int_{0}^{a-t}{\sqrt{a^{2}-t^{2}}}dt
Now apply the formula and get ur answer[1]