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Definite Integration question

2020 · Shift 1 · Q30
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Definite Integration question

2020 · Shift 1 · Q30

JEE AdvancedMathematicsDefinite IntegrationMultiple correct+4 / −2
Which of the following inequalities is/are TRUE?
  1. A
    ∫01xcos⁡xdx ≥ 38\int_0^1 {x\cos xdx\, \ge \,{3 \over 8}}∫01​xcosxdx≥83​
  2. B
    ∫01xsin⁡xdx ≥ 310\int_0^1 {x\sin xdx\, \ge \,{3 \over {10}}}∫01​xsinxdx≥103​
  3. C
    ∫01x2cos⁡xdx ≥ 12\int_0^1 {{x^2}\cos xdx\, \ge \,{1 \over 2}}∫01​x2cosxdx≥21​
  4. D
    ∫01x2sin⁡xdx ≥ 29\int_0^1 {{x^2}\sin xdx\, \ge \,{2 \over 9}}∫01​x2sinxdx≥92​
View written solutionFree

Correct answer: A, B, D

We evaluate each integral exactly and compare with the given bound.

1. Option A

We need to check

∫01xcos⁡x dx≥38.\int_0^1 x\cos x\,dx \ge \frac{3}{8}.∫01​xcosxdx≥83​.

Using integration by parts:

∫xcos⁡x dx=xsin⁡x−∫sin⁡x dx=xsin⁡x+cos⁡x.\int x\cos x\,dx = x\sin x-\int \sin x\,dx = x\sin x+\cos x.∫xcosxdx=xsinx−∫sinxdx=xsinx+cosx.

So,

∫01xcos⁡x dx=[xsin⁡x+cos⁡x]01=(sin⁡1+cos⁡1)−1.\int_0^1 x\cos x\,dx = \left[x\sin x+\cos x\right]_0^1 = (\sin 1+\cos 1)-1.∫01​xcosxdx=[xsinx+cosx]01​=(sin1+cos1)−1.

Now use standard values:

sin⁡1≈0.84147,cos⁡1≈0.54030.\sin 1\approx 0.84147, \qquad \cos 1\approx 0.54030.sin1≈0.84147,cos1≈0.54030.

Hence

∫01xcos⁡x dx≈0.84147+0.54030−1=0.38177.\int_0^1 x\cos x\,dx \approx 0.84147+0.54030-1=0.38177.∫01​xcosxdx≈0.84147+0.54030−1=0.38177.

Also,

38=0.375.\frac{3}{8}=0.375.83​=0.375.

Thus

0.38177>0.375.0.38177>0.375.0.38177>0.375.

So A is true.

2. Option B

We need to check

∫01xsin⁡x dx≥310.\int_0^1 x\sin x\,dx \ge \frac{3}{10}.∫01​xsinxdx≥103​.

Using integration by parts:

∫xsin⁡x dx=−xcos⁡x+∫cos⁡x dx=−xcos⁡x+sin⁡x.\int x\sin x\,dx = -x\cos x+\int \cos x\,dx = -x\cos x+\sin x.∫xsinxdx=−xcosx+∫cosxdx=−xcosx+sinx.

Therefore,

∫01xsin⁡x dx=[−xcos⁡x+sin⁡x]01=sin⁡1−cos⁡1.\int_0^1 x\sin x\,dx = \left[-x\cos x+\sin x\right]_0^1 = \sin 1-\cos 1.∫01​xsinxdx=[−xcosx+sinx]01​=sin1−cos1.

Numerically,

sin⁡1−cos⁡1≈0.84147−0.54030=0.30117.\sin 1-\cos 1 \approx 0.84147-0.54030=0.30117.sin1−cos1≈0.84147−0.54030=0.30117.

And

310=0.3.\frac{3}{10}=0.3.103​=0.3.

Thus

0.30117>0.3.0.30117>0.3.0.30117>0.3.

So B is true.

3. Option C

We need to check

∫01x2cos⁡x dx≥12.\int_0^1 x^2\cos x\,dx \ge \frac{1}{2}.∫01​x2cosxdx≥21​.

Use integration by parts:

I=∫x2cos⁡x dx.I=\int x^2\cos x\,dx.I=∫x2cosxdx.

Take

u=x2,dw=cos⁡x dx.u=x^2,\quad dw=\cos x\,dx.u=x2,dw=cosxdx.

Then

du=2x dx,w=sin⁡x.du=2x\,dx,\quad w=\sin x.du=2xdx,w=sinx.

So

I=x2sin⁡x−2∫xsin⁡x dx.I=x^2\sin x-2\int x\sin x\,dx.I=x2sinx−2∫xsinxdx.

We already know

∫xsin⁡x dx=−xcos⁡x+sin⁡x.\int x\sin x\,dx=-x\cos x+\sin x.∫xsinxdx=−xcosx+sinx.

Hence

I=x2sin⁡x−2(−xcos⁡x+sin⁡x)=x2sin⁡x+2xcos⁡x−2sin⁡x.I=x^2\sin x-2(-x\cos x+\sin x) = x^2\sin x+2x\cos x-2\sin x.I=x2sinx−2(−xcosx+sinx)=x2sinx+2xcosx−2sinx.

Therefore,

∫01x2cos⁡x dx=[x2sin⁡x+2xcos⁡x−2sin⁡x]01=(sin⁡1+2cos⁡1−2sin⁡1)−0=2cos⁡1−sin⁡1.\int_0^1 x^2\cos x\,dx =\left[x^2\sin x+2x\cos x-2\sin x\right]_0^1 = (\sin 1+2\cos 1-2\sin 1)-0 =2\cos 1-\sin 1.∫01​x2cosxdx=[x2sinx+2xcosx−2sinx]01​=(sin1+2cos1−2sin1)−0=2cos1−sin1.

Numerically,

2cos⁡1−sin⁡1≈2(0.54030)−0.84147=0.23913.2\cos 1-\sin 1 \approx 2(0.54030)-0.84147=0.23913.2cos1−sin1≈2(0.54030)−0.84147=0.23913.

But

12=0.5.\frac{1}{2}=0.5.21​=0.5.

Thus

0.23913<0.5.0.23913<0.5.0.23913<0.5.

So C is false.

4. Option D

We need to check

∫01x2sin⁡x dx≥29.\int_0^1 x^2\sin x\,dx \ge \frac{2}{9}.∫01​x2sinxdx≥92​.

Let

J=∫x2sin⁡x dx.J=\int x^2\sin x\,dx.J=∫x2sinxdx.

Using integration by parts:

u=x2,dw=sin⁡x dxu=x^2,\quad dw=\sin x\,dxu=x2,dw=sinxdx

so that

du=2x dx,w=−cos⁡x.du=2x\,dx,\quad w=-\cos x.du=2xdx,w=−cosx.

Then

J=−x2cos⁡x+2∫xcos⁡x dx.J=-x^2\cos x+2\int x\cos x\,dx.J=−x2cosx+2∫xcosxdx.

Now,

∫xcos⁡x dx=xsin⁡x+cos⁡x.\int x\cos x\,dx = x\sin x+\cos x.∫xcosxdx=xsinx+cosx.

Hence

J=−x2cos⁡x+2(xsin⁡x+cos⁡x).J=-x^2\cos x+2(x\sin x+\cos x).J=−x2cosx+2(xsinx+cosx).

Therefore,

∫01x2sin⁡x dx=[−x2cos⁡x+2xsin⁡x+2cos⁡x]01\int_0^1 x^2\sin x\,dx =\left[-x^2\cos x+2x\sin x+2\cos x\right]_0^1∫01​x2sinxdx=[−x2cosx+2xsinx+2cosx]01​ =(−cos⁡1+2sin⁡1+2cos⁡1)−2=cos⁡1+2sin⁡1−2.=(-\cos 1+2\sin 1+2\cos 1)-2 =\cos 1+2\sin 1-2.=(−cos1+2sin1+2cos1)−2=cos1+2sin1−2.

Numerically,

cos⁡1+2sin⁡1−2≈0.54030+2(0.84147)−2=0.22324.\cos 1+2\sin 1-2 \approx 0.54030+2(0.84147)-2=0.22324.cos1+2sin1−2≈0.54030+2(0.84147)−2=0.22324.

Now,

29=0.22222…\frac{2}{9}=0.22222\ldots92​=0.22222…

Thus

0.22324>0.22222…0.22324>0.22222\ldots0.22324>0.22222…

So D is true.

5. Final conclusion

The true inequalities are:

A, B, D\boxed{A,\ B,\ D}A, B, D​

This matches the stored correct answer.

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