Menu Close

Find-0-pi-f-x-cosxdx-given-that-f-x-1-2-k-1-n-coskx-




Question Number 1780 by 112358 last updated on 23/Sep/15
Find                      ∫_0 ^( π) f(x)cosxdx  given that                  f(x)=1+2Σ_(k=1) ^n coskx.
Find0πf(x)cosxdxgiventhatf(x)=1+2nk=1coskx.
Commented by Rasheed Soomro last updated on 23/Sep/15
      ∫_0 ^( π) f(x)cosxdx  f(x)=1+2Σ_(k=1) ^n coskx.   −−−−−−−−−−−−−−−  ∫_0 ^( π) (1+2Σ_(k=1) ^n coskx)cosxdx  ∫_0 ^( π) cosxdx+2Σ_(k=1) ^n ∫_0 ^( π) (coskx)(cosx)dx
0πf(x)cosxdxf(x)=1+2nk=1coskx.0π(1+2nk=1coskx)cosxdx0πcosxdx+2nk=10π(coskx)(cosx)dx
Answered by 123456 last updated on 24/Sep/15
f(x)=1+2Σ_(k=1) ^n cos kx  f(x)cos x=cos x+2Σ_(k=1) ^n cos kx cos x  ∫_0 ^π f(x)cos x dx=∫_0 ^π cos x+2Σ_(k=1) ^n ∫_0 ^π cos kx cos xdx  ∫_0 ^π cos xdx=sin π−sin 0=0   ∫_0 ^π cos^2 xdx=(1/2)∫_0 ^π dx+(1/2)∫_0 ^π cos 2xdx=(1/2)π  continue
f(x)=1+2nk=1coskxf(x)cosx=cosx+2nk=1coskxcosxπ0f(x)cosxdx=π0cosx+2nk=1π0coskxcosxdxπ0cosxdx=sinπsin0=0π0cos2xdx=12π0dx+12π0cos2xdx=12πcontinue
Commented by 112358 last updated on 29/Sep/15
Graphically I found the value  of the integral to be approximately π.  Consider the identity  coskxcosx=(1/2)(cos(k+1)x+cos(k−1)x)  Then the integral under the   summation becomes  ∫_0 ^π (cos(k+1)x+cos(k−1)x)dx  =(1/(k+1))sin(k+1)x+(1/(k−1))sin(k−1)x∣_0 ^π   =((sin(k+1)π)/(k+1))+((sin(k−1)π)/(k−1))  With k∈Z^+  and k≥1 , sin(k+1)π=0  and sin(k−1)π=0. But  ((sin(k−1)π)/(k−1))  is undefined at k=1.  ∴ ∫_0 ^( π) f(x)cosx dx=πΣ_(k=1) ^n ((sin(k−1)π)/((k−1)π))  If k were a continuous variable  we get that lim_(k→1) ((sin(k−1)π)/((k−1)π))=1  so that ∫_0 ^π f(x)cosxdx=π  but I don′t feel comfortable with  this approach since k is a discrete  quantity.
GraphicallyIfoundthevalueoftheintegraltobeapproximatelyπ.Considertheidentitycoskxcosx=12(cos(k+1)x+cos(k1)x)Thentheintegralunderthesummationbecomes0π(cos(k+1)x+cos(k1)x)dx=1k+1sin(k+1)x+1k1sin(k1)x0π=sin(k+1)πk+1+sin(k1)πk1WithkZ+andk1,sin(k+1)π=0andsin(k1)π=0.Butsin(k1)πk1isundefinedatk=1.0πf(x)cosxdx=πnk=1sin(k1)π(k1)πIfkwereacontinuousvariablewegetthatlimk1sin(k1)π(k1)π=1sothat0πf(x)cosxdx=πbutIdontfeelcomfortablewiththisapproachsincekisadiscretequantity.

Leave a Reply

Your email address will not be published. Required fields are marked *