If I is the greatest of the definite integrals I 1-011 e - x cos 2 x dx - I 2-01 e - x 2cos 2 xdxI 3-01 e - x 2 dx - I 4-01 e - x 2 - 2 dx- thenA- I - I 1B- I - I 2C- I - I 3D- I - I 4

The correct option is D I = I_4For (0 lt; x lt; 1, we have 1/2x^2 lt; x^2 lt; x ⇒ x^2 gt; x, so that e^ x^2 lt; e^ x, Hence ∫_0^1 e^ x^2 co ...

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Byju's Answer

Standard XII

Mathematics

Geometric Interpretation of Def.Int as Limit of Sum

If I is the g...

Question

If $I$ is the greatest of the definite integrals
$I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x, I_{2} \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x$
$I_{3} = \int_{0}^{1} e^{- x^{2}} d x, I_{4} = \int_{0}^{1} e^{- x^{2} / 2} d x,$ then

I = I_{1}

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I = I_{2}

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I = I_{3}

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I = I_{4}

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Solution

The correct option is D $I = I_{4}$
For (0 < x < 1, we have $\frac{1}{2} x^{2} < x^{2} < x$
$\Rightarrow - x^{2} > - x,$ so that $e^{- x^{2}} < e^{- x},$
Hence $\int_{0}^{1} e^{- x^{2}} c o s^{2} x d x > \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x .$
Also $c o s^{2} x \leq 1$
$T h e r e f o r e \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x \leq \int_{0}^{1} e^{- x^{2}} d x < \int_{0}^{1} e^{- x^{2} / 2} d x = I_{4}$
Hence $I_{4}$ is the greatest integral

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Similar questions

If I is the greatest of the definite integrals
$I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x, I_{2} = \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x$
$I_{3} = \int_{0}^{1} e^{- x^{2}} d x, I_{4} = \int_{0}^{1} e^{\frac{- x^{2}}{2}} d x, t h e n$

Q. Consider the integrals

I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x

I_{2} = \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x, I_{3} = \int_{0}^{1} e^{- x^{2}} d x

and

I_{4} = \int_{0}^{1} e^{- (1 / 2) x^{2}} d x

. The greatest of these integrals

I

Q. Assertion :Let

I_{1} = \int_{0}^{1} e^{- x} {cos}^{2} x d x

I_{2} = \int_{0}^{1} e^{- x^{2}} {cos}^{2} x d x

I_{3} = \int_{0}^{1} e^{- x^{2}} d x

I_{4} = \int_{0}^{1} e^{\frac{- x^{2}}{2}} d x

then greatest of these integrals is

I_{4}

Reason:

\int_{a}^{b} f (x) d x < \int_{a}^{b} g (x) \forall f (x) \geq g (x)

Consider the integrals $I_{1} = \int_{0}^{1} e^{- x} {cos}^{2} x d x, I_{2} = \int_{0}^{1} e^{- x^{2}} {cos}^{2} x d x, I_{3} = \int_{0}^{1} e^{- x} d x$ and $I_{4} = \int_{0}^{1} e^{- (1 / 2) x^{2}} d x$ . The greatest of these integrals is

Q. Consider the integrals

I_{1} = 1 \int 0 e^{- x} {cos}^{2} x d x, I_{2} = 1 \int 0 e^{- x^{2}} cos x d x

I_{3} = 1 \int 0 e^{- x^{2}} d x and I_{4} = 1 \int 0 e^{- x^{2} / 2} d x .

Which of the following is/are correct?

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If I is the greatest of the definite integrals I1=∫10e−xcos2 x dx,I2∫10e−x2 cos2 xdx I3=∫10e−x2 dx,I4=∫10e−x2/2dx, then

The correct option is D I=I4For (0 < x < 1, we have 12x2<x2<x ⇒−x2>−x, so that e−x2<e−x, Hence ∫10e−x2 cos2 xdx>∫10e−x2 cos2 x dx. Also cos2x≤1 Therefore∫10e−x2cos2 xdx≤∫10e−x2dx<∫10e−x2/2dx=I4 Hence I4 is the greatest integral

If $I$ is the greatest of the definite integrals
$I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x, I_{2} \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x$
$I_{3} = \int_{0}^{1} e^{- x^{2}} d x, I_{4} = \int_{0}^{1} e^{- x^{2} / 2} d x,$ then