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0-100-sin-x-x-pi-dx-




Question Number 83225 by 09658867628 last updated on 28/Feb/20
 ∫_( 0) ^(100)  sin (x−[x])π dx =
1000sin(x[x])πdx=
Commented by mathmax by abdo last updated on 29/Feb/20
A =Σ_(n=0) ^(99) ∫_n ^(n+1) sin((x−n)π)dx  =Σ_(n=0) ^(99)  ∫_n ^(n+1) sin(πx−nπ)dx =Σ_(n=0) ^(99)  ∫_n ^(n+1) (−1)^n sin(πx)dx  =Σ_(n=0) ^(99) (−1)^n   [−(1/π)cos(πx)]_n ^(n+1)   =(1/π)Σ_(n=0) ^(99) (−1)^(n+1) {cos(n+1)π −cos(nπ)}  =(1/π)Σ_(n=0) ^(99) (−1)^(n+1) {(−1)^(n+1) −(−1)^n }  =−(2/π)Σ_(n=0) ^(99) (−1)^n ×(−1)^(n+1)  =(2/π)Σ_(n=0) ^(99) (1)  =((200)/π)
A=n=099nn+1sin((xn)π)dx=n=099nn+1sin(πxnπ)dx=n=099nn+1(1)nsin(πx)dx=n=099(1)n[1πcos(πx)]nn+1=1πn=099(1)n+1{cos(n+1)πcos(nπ)}=1πn=099(1)n+1{(1)n+1(1)n}=2πn=099(1)n×(1)n+1=2πn=099(1)=200π

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