The value of I - -2-2-sin x- dx is

The explanation for the correct answer.Evaluate the integral.I = ∫_ π/2^π/2|sin x| dx|sin x|is even function.∴ I=2∫_0^π/2sin x dx0ex0ex=2[( cos x )]_0^π/20ex0e ...

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Standard XI

Mathematics

Rationalization Method to Remove Indeterminate Form

The value of ...

Question

The value of $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} | \sin x | d x$ is

$0$

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$2$

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$- 2$

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$- 2 < I < 2$

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Solution

The correct option is B
$2$

The explanation for the correct answer.
Evaluate the integral.
$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} | \sin x | d x$
$|\sin x|$ is even function.
$∴ I = 2 \int_{0}^{\frac{π}{2}} \sin x d x = 2 {[(- \cos x)]}_{0}^{\frac{π}{2}} = 2$
Hence, option B is correct .

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The value of I=∫-π2π2|sinx|dx is

The correct option is B 2The explanation for the correct answer.Evaluate the integral.I=∫-π2π2|sinx|dxsinxis even function.∴I=2∫0π2sinxdx=2-cosx0π2=2Hence, option B is correct .

The value of $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} | \sin x | d x$ is

The correct option is B
$2$

The explanation for the correct answer.
Evaluate the integral.
$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} | \sin x | d x$
$|\sin x|$ is even function.
$∴ I = 2 \int_{0}^{\frac{π}{2}} \sin x d x = 2 {[(- \cos x)]}_{0}^{\frac{π}{2}} = 2$
Hence, option B is correct .