The value of -n-110 -2n-1-2nsin27x dx-n-110-2n2n-1sin27x dx-

∑_n=1^10 ∫_ 2n 1^ 2nsin^27x dx+∑_n=1^10∫_2n^2n+1sin^27xdx [in the first summation put x = t] =∑_n=1^10 ∫_2n+1^2nsin^27( t)( dt)+∑_n=1^10 ∫_2n^2n+1sin^27x dx ...

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Standard XII

Mathematics

Integration of Piecewise Continuous Functions

The value of ...

Question

The value of $10 \sum n = 1 - 2 n \int - 2 n - 1 {sin}^{27} x d x + 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x =$

27

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Solution

The correct option is D $0$
$10 \sum n = 1 - 2 n \int - 2 n - 1 {sin}^{27} x d x + 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x$

[in the first summation put $x = - t$ ]

$= 10 \sum n = 1 2 n \int 2 n + 1 {sin}^{27} (- t) (- d t) + 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x$

$= 10 \sum n = 1 2 n \int 2 n + 1 {sin}^{27} x d x + 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x = 0$

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Integration of Piecewise Continuous Functions

Standard XII Mathematics

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The value of 10∑n=1 −2n∫−2n−1sin27x dx+10∑n=12n+1∫2nsin27x dx=

The correct option is D 010∑n=1 −2n∫−2n−1sin27xdx+10∑n=12n+1∫2nsin27xdx [in the first summation put x=−t] =10∑n=1 2n∫2n+1sin27(−t)(−dt)+10∑n=1 2n+1∫2nsin27x dx =10∑n=1 2n∫2n+1sin27x dx+10∑n=1 2n+1∫2nsin27x dx=0

The value of $10 \sum n = 1 - 2 n \int - 2 n - 1 {sin}^{27} x d x + 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x =$