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Q) If for a real number y,[y] is the greatest integer less than or equal to y, then find the value of the integral $\int^{\frac{3\pi}{2}}_{\frac{\pi}{2}}\left[2sinx\right]dx$.

Attempt : enter image description here

On adding up individual areas I'm getting sum = $\frac{7\pi}{6}$ whereas the answer is $\frac{-\pi}{2}$ . Where am i wrong ?

Im getting total area (taking modulus for -ve area) = $\frac{7\pi}{6}$ But without considering modulus of -ve part and just adding with im getting it correct $\frac{-\pi}{2}$ Im not able to understand this , will definite integral ever yield a -ve area ?

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When $sin(x)>0$ we get $f=0$ whenever $0<x<\frac{\pi}{6}$ and $\frac{5\pi}{6}<x<\pi$.
But we get $f=1$ between $\frac{\pi}{6}<x<\frac{5\pi}{6}$. Since this interval is excluded, we have $f=1$ Whenever $sin(x)>0$

Do the similar partition for $sin(x)<0$

Use limit since there are some jump points. Calculate again the result.

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  • $\begingroup$ Okay but for $sinx<0$ shouldn't i take the modulus of the area and add it with with the rest . $\endgroup$ – Sujith Sizon Sep 19 '15 at 9:21
  • $\begingroup$ m getting total area (taking modulus for -ve area) = $\frac{7\pi}{6}$ But without considering modulus of -ve part and just adding with im getting it correct $\frac{-\pi}{2} Im not able to understand this , will definite integral ever yield a -ve area ? $\endgroup$ – Sujith Sizon Sep 19 '15 at 9:35
  • $\begingroup$ Why are you taking modulus? Are you trying to calculate integral or are you trying calculate area by help of integral? $\endgroup$ – Salihcyilmaz Sep 19 '15 at 11:35
  • $\begingroup$ Integral only . so we never take modulus while solving definite integral ? $\endgroup$ – Sujith Sizon Sep 19 '15 at 12:01
  • $\begingroup$ aargh i confused over area under curve and integration , sorry . $\endgroup$ – Sujith Sizon Sep 19 '15 at 12:03

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