If I 1-01 e - x cos 2 x dx-I 2-01 e -x2cos 2 x dx andI 3-01 e - x 3 dx - then -A- I 3- I 2- I 1B- I 2- I 1- I 3C- I 2- I 3- I 1D- I 3- I 1- I 2

As x∈ (0,1) So, x gt;x^2 gt;x^3 ⇒ x^3 x^2 x ⇒ e^ x^3 e^ x^2 e^ x ⇒ e^ x^3cos^2x e^ x^2cos^2x e^ x ∵cos^2x 1 ∀ x∈ (0,1) ⇒∫_0^1e^ x^3 dx∫_0^1e^ x^2c ...

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Byju's Answer

Standard XII

Mathematics

Properties of Inequalities

If I 1=∫01 e ...

Question

If $I_{1} = 1 \int 0 e^{- x} {cos}^{2} x d x$ ,
$I_{2} = 1 \int 0 e^{- x^{2}} {cos}^{2} x d x$ and
$I_{3} = 1 \int 0 e^{- x^{3}} d x$ ; then :

I_{3} > I_{2} > I_{1}

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I_{2} > I_{1} > I_{3}

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I_{2} > I_{3} > I_{1}

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I_{3} > I_{1} > I_{2}

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Solution

The correct option is A $I_{3} > I_{2} > I_{1}$
As $x \in (0, 1)$
So, $x > x^{2} > x^{3}$
$\Rightarrow - x^{3} < - x^{2} < - x$
$\Rightarrow e^{- x^{3}} > e^{- x^{2}} > e^{- x}$
$\Rightarrow e^{- x^{3}} > {cos}^{2} x e^{- x^{2}} > {cos}^{2} x e^{- x}$
$∵ {cos}^{2} x < 1 \forall x \in (0, 1)$
$\Rightarrow 1 \int 0 e^{- x^{3}} d x > 1 \int 0 e^{- x^{2}} {cos}^{2} x d x > 1 \int 0 e^{- x} {cos}^{2} x d x$
$∴ I_{3} > I_{2} > I_{1}$

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Similar questions

Q. If

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^{3}} d x; then :

Q. Let

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

and

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

Then

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)

Q. If

I_{1} = π / 2 \int 0 cos (sin x) d x; I_{2} = π / 2 \int 0 sin (cos x) d x

and

I_{3} = π / 2 \int 0 cos x d x,

then

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If I1=1∫0e−xcos2x dx, I2=1∫0e−x2cos2x dx and I3=1∫0e−x3 dx ; then :

The correct option is A I3>I2>I1As x∈(0,1) So, x>x2>x3 ⇒−x3<−x2<−x ⇒e−x3>e−x2>e−x ⇒e−x3>cos2x e−x2>cos2x e−x ∵cos2x<1 ∀ x∈(0,1) ⇒1∫0e−x3 dx>1∫0e−x2cos2x dx>1∫0e−xcos2x dx ∴I3>I2>I1

If $I_{1} = 1 \int 0 e^{- x} {cos}^{2} x d x$ ,
$I_{2} = 1 \int 0 e^{- x^{2}} {cos}^{2} x d x$ and
$I_{3} = 1 \int 0 e^{- x^{3}} d x$ ; then :