Question

integral 0 to 10 π modulus of sin x dx

Answer 1

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Answer 2

$\int\limits^{(10\pi)}_b {[\sin x]} \, dx=20$

Step-by-step explanation:

We have,

$\int\limits^{(10\pi)}_b {[\sin x]} \, dx$

To find, $\int\limits^{(10\pi)}_b {[\sin x]} \, dx=?$

We know that,

$\sin(x+\pi)=-sin(x)$

In any periodic function with period T,

$\int\limits^{(a+kT)}_a {f(x)} \, dx =k\int\limits^{(a+T)}_a {f(x)} \, dx$

∴ $\int\limits^{(10\pi)}_b {[\sin x]} \, dx=10\int\limits^{(\pi)}_b {[\sin x]} \, dx$

$=10[-\cos x]_{0}^{\pi}$

$=10[\cos 0-\cos \pi]$

$=10[1-(-1)]$

$=10[1+1]$

=  20

Hence, $\int\limits^{(10\pi)}_b {[\sin x]} \, dx=20.$