Evaluate using Substitution MCQ Quiz - Objective Question with Answer for Evaluate using Substitution - Download Free PDF

Explanation:

Evaluation of Integrals \(I_1\) and \(I_2\)

We are given two integrals:

\(I_{1}=\int_{0}^{1} \frac{\operatorname{cosec} x}{x} dx\)

\(I_{2}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{m} x}{x^{n}} dx\)

We need to determine the conditions under which these integrals converge or diverge.

Analysis of \(I_1\):

The integral \(I_1\) is given by:

\(I_{1}=\int_{0}^{1} \frac{\operatorname{cosec} x}{x} dx\)

To analyze the convergence of this integral, we need to consider the behavior of the integrand near the limits of integration, particularly near \(x = 0\).

As \(x \to 0\), \(\operatorname{cosec} x = \frac{1}{\sin x}\) behaves like \(\frac{1}{x}\). Therefore, near \(x = 0\), the integrand \(\frac{\operatorname{cosec} x}{x}\) behaves like \(\frac{1}{x^2}\).

The integral \(\int_{0}^{\epsilon} \frac{1}{x^2} dx\) diverges as \(\epsilon \to 0\). Thus, \(I_1\) diverges.

Analysis of \(I_2\):

The integral \(I_2\) is given by:

\(I_{2}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{m} x}{x^{n}} dx\)

To determine the conditions for convergence of \(I_2\), we consider the behavior of the integrand near \(x = 0\) and \(x = \frac{\pi}{2}\).

Near \(x = \frac{\pi}{2}\), \(\sin x\) is close to 1, so the integrand does not pose any convergence issues. The critical part to analyze is the behavior near \(x = 0\).

As \(x \to 0\), \(\sin x \approx x\). Therefore, near \(x = 0\), the integrand \(\frac{\sin ^{m} x}{x^{n}}\) behaves like \(\frac{x^m}{x^n} = x^{m-n}\).

The integral \(\int_{0}^{\epsilon} x^{m-n} dx\) converges if and only if \(m-n > -1\), which simplifies to \(n < m + 1\).

Therefore, \(I_2\) converges if \(n < m + 1\).

Conclusion:

Based on the above analysis, we can conclude that:

\(I_1\) diverges and \(I_2\) converges if \(n < m + 1\).

Therefore, the correct option is:

Option 4: \(I_1\) diverges and \(I_2\) converges if \(n < m + 1\).

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: \(I_1\) converges and \(I_2\) converges if \(n \geq m + 1\).

This option is incorrect because \(I_1\) diverges as shown in the analysis. \(I_2\) converges when \(n < m + 1\), not \(n \geq m + 1\).

Option 2: \(I_1\) converges and \(I_2\) converges if \(n, m \in \mathbb{N}\).

This option is incorrect because \(I_1\) diverges regardless of the values of \(n\) and \(m\).

Option 3: \(I_1\) diverges and \(I_2\) diverges if \(n < m 1\).

This option is incorrect because \(I_2\) converges if \(n < m + 1\), which is the opposite condition to that stated.

Understanding the convergence properties of these integrals helps in correctly identifying the conditions under which they converge or diverge. This is crucial for solving problems in advanced calculus and mathematical analysis.

Calculation

For I

Apply king (P– 5) and add 

⇒ \(2 \mathrm{I}=\int_{0}^{\pi / 2} \mathrm{dx}=\frac{\pi}{2} \Rightarrow \mathrm{I}=\frac{\pi}{4}\)

\(I_{2}=\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x\)

Apply king and add 

⇒ \(\mathrm{I}_{2}=\frac{\pi}{4} \int_{0}^{\pi / 2} \frac{\tan \mathrm{x} \sec ^{2} \mathrm{xdx}}{\tan ^{4} \mathrm{x}+1}\)

put tan² x = t 

⇒ \(\frac{\pi}{8} \int_{0}^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}\)

⇒ \(\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^{2}}{16}\)

Hence option 1 is correct

Calculation 

f(x) = (7tan⁶ x – 3tan² x)(sec² x)

\(I_{1}=\int_{0}^{\pi / 4}\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right) d x\)

Put tanx = t

\(I_{1}=\int_{0}^{1}\left(7 t^{6}-3 t^{2}\right) d t=\left[t^{7}-t^{3}\right]_{0}^{1}=0\)

\(I_{2}=\int_{0}^{\pi / 4} x \underbrace{\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right)}_{\mathrm{II}} d x\)

= \(\left[x\left(\tan ^{7} x-\tan ^{3} x\right)\right]_{0}^{\pi / 4}-\int_{0}^{\pi / 4}\left(\tan ^{7} x-\tan ^{3} x\right) d x\)

= \(0-\int_{0}^{\pi / 4} \tan ^{3} x\left(\tan ^{2} x-1\right)\left(1+\tan ^{2} x\right) d x\)

Put tanx = t

= \(-\int_{0}^{1}\left(t^{5}-t^{3}\right) d t=-\left[\frac{t^{6}}{6}-\frac{t^{4}}{4}\right]=\frac{1}{12}\)

7I₁ + 12I₂ = 1 

Hence option 3 is correct

Calculation

Given: \(\int_0^1 (0.001)^{\frac{x}{3}} \cdot e^x dx\)

I = \(\int_0^1 (10^{-3})^{\frac{x}{3}} \cdot e^x dx = \int_0^1 10^{-x} \cdot e^x dx\)

⇒ I = \(\int_0^1 e^{-x \ln 10} \cdot e^x dx = \int_0^1 e^{x(1 - \ln 10)} dx\)

Let \(u = x(1 - \ln 10)\). Then \(\frac{du}{dx} = 1 - \ln 10\), so \(du = (1 - \ln 10) dx\).

When \(x = 0\), \(u = 0\). When \(x = 1\), \(u = 1 - \ln 10\).

I = \(\int_0^{1 - \ln 10} e^u \frac{du}{1 - \ln 10} = \frac{1}{1 - \ln 10} \int_0^{1 - \ln 10} e^u du\)

⇒ I =  \(\frac{1}{1 - \ln 10} [e^u]_0^{1 - \ln 10} = \frac{1}{1 - \ln 10} (e^{1 - \ln 10} - e^0)\)

⇒ I = \(\frac{1}{1 - \ln 10} (e^{1 - \ln 10} - 1) = \frac{e^{1 - \ln 10} - 1}{1 - \ln 10}\)

⇒ I = \(\frac{e \cdot e^{-\ln 10} - 1}{1 - \ln 10} = \frac{e \cdot \frac{1}{10} - 1}{1 - \ln 10} = \frac{\frac{e}{10} - 1}{1 - \ln 10}\)

⇒ I = \(\frac{\frac{e - 10}{10}}{1 - \ln 10} = \frac{e - 10}{10(1 - \ln 10)}\)

Hence option 3 is correct.

Calculation

Given \(I_n = \int_{0}^{\pi/4} \tan^n x \, dx\)

\(I_n = \int_{0}^{\pi/4} \tan^n x \, dx = \int_{0}^{\pi/4} \tan^{n-2} x \tan^2 x \, dx\)

\(= \int_{0}^{\pi/4} \tan^{n-2} x (\sec^2 x - 1) \, dx\)

\(= \int_{0}^{\pi/4} \tan^{n-2} x \sec^2 x \, dx - \int_{0}^{\pi/4} \tan^{n-2} x \, dx\)

\(= \int_{0}^{\pi/4} \tan^{n-2} x \sec^2 x \, dx - I_{n-2}\)

Let \(u = \tan x\), then \(du = \sec^2 x \, dx\).

When \(x = 0\), \(u = 0\) and when \(x = \pi/4\), \(u = 1\).

\(\int_{0}^{\pi/4} \tan^{n-2} x \sec^2 x \, dx = \int_{0}^{1} u^{n-2} \, du = \left[ \frac{u^{n-1}}{n-1} \right]_{0}^{1} = \frac{1}{n-1}\)

Therefore, \(I_n = \frac{1}{n-1} - I_{n-2}\)

Or, \(I_n + I_{n-2} = \frac{1}{n-1}\)

We need to find \(I_{13} + I_{11}\). Using the formula above, we have:

\(I_{13} + I_{11} = \frac{1}{13-1} = \frac{1}{12}\)

Hence option 2 is correct

Concept:

\(\rm \dfrac{d(\ln x)}{dx} = \dfrac{1}{x}\)

Calculation:

Let I = \(\rm \int_{1}^{e}\frac{dx}{x\sqrt{2+\ln x}}\)

Let (2 + ln x) = t²

Differentiating with respect to x, we get

⇒ (0 + \(\rm \frac 1 x\))dx = 2tdt

⇒ \(\rm \frac 1 x\)dx = 2tdt

Now,

\(\rm I=\rm \int_{\sqrt 2}^{\sqrt 3}\frac{2tdt}{\sqrt{t^2}}\\=2\int_{\sqrt 2}^{\sqrt 3}\frac{tdt}{t}\\=2\int_{\sqrt 2}^{\sqrt 3}dt\\=2[t]_{\sqrt 2}^{\sqrt 3}\\=2(\sqrt 3- \sqrt 2)\)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C

Calculation:

I = \(\rm\int {2x+1\over\left({x^2+x+1}\right)^2}dx\)

Let x2 + x + 1 = t

⇒ (2x + 1) dx = dt

I = \(\rm \int {dt\over t^2}\)

I = \(\rm {t^{-1}\over{-1}}\)

I = \(\rm -1\over t\)

I = \(\rm -1\over x^2+x+1\)

Putting limits

I = \(\rm \left[-1\over x^2+x+1\right]_{-1}^1\)

I = \(\rm {-1\over 1^2+1+1} - \left({-1\over (-1)^2+(-1)+1}\right)\)

I = \(\rm 1-{1\over3}\) = \(\boldsymbol{2\over3}\)

Concept:

\( \rm \int x^{n}dx = \frac{x^{n+1}}{n+1} + c\)

Calculation:

Let I = \(\rm \int^1_0 \frac {\tan^{-1} x}{1 + x^2}dx\)

Let tan^-1 x = t

Differentiating both sides, we get

⇒ \(\rm \frac{1}{1+x^2} dx = dt\) 

Now, I \(= \rm \int _0^{\pi/4} t\;dt\)

\( = \rm [\frac{t^2}{2}]_0^{\pi/4}\)

\(= \rm \frac{1}{2}[\frac{\pi^2}{16}-0]\)

= \(\rm \frac {\pi^2}{32}\)

Concept:

\(\rm \int x^n dx=\frac{x^{n+1}}{n+1}+c\)

Calculation:

Let I = \(\rm \int_{0}^{1}{e}^{{x}}({e}^{{x}}-1)^{4}dx\)

Let (e^x - 1) = t

Differentiating with respect to x, we get

⇒ e^xdx = dt

\(\rm I = \int_{0}^{1} t^{4}dt \\=\left[\frac{t^{5}}{5}\right]_0^1\)

Put t = (e^x - 1) 

\(\rm =\left[\frac{(e^x-1)^{5}}{5}\right]_0^1\\=\left[\frac{(e^1-1)^{5}}{5} -\frac{(e^0-1)^{5}}{5} \right]\\=\frac{(e-1)^{5}}{5}\)

Concept:

• ∫ sin x dx = -cos x + C
• ∫ cos x dx = sin x + C
• ∫ xⁿ dx = \(\rm {x^{n+1}\over n+1}\) + C

Calculation:

I = \(\rm ∫\sin^2x\cos x\)dx

Let sin x = t

dt = cos x dx

⇒ I = ∫ t² (dt)

⇒ I =  \(\rm t^3\over3\) + c

⇒ I = \(\rm {\sin^3x\over3}\) + c

Putting the limits 

⇒ I = \(\rm \left[{\sin^3x\over3}+c\right]^{\pi\over2}_0\)

⇒ I = \(\rm \left[{\sin^3{\pi\over2}\over3}+c\right]-\left[{\sin^3(0)\over3}+c\right]\)

⇒ I = \(\boldsymbol {\rm {1\over3}}\)

Concept:

\(\smallint \frac{1}{{1 + {{\rm{x}}^2}}}{\rm{dx}} = {\tan ^{ - 1}}{\rm{x}} + {\rm{c}}\)

Calculation:

I = \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\cos {\rm{x}}}}{{1 + {{\sin }^2}{\rm{x}}}}{\rm{dx}}\)

Let sin x = t

⇒ cos x dx = dt

\({\rm{I}} = \mathop \smallint \nolimits_0^1 \frac{1}{{1 + {{\rm{t}}^2}}}{\rm{dt}} \)

\(= {\rm{\;}}\left[ {{{\tan }^{ - 1}}{\rm{t}}} \right]_0^1\)

= tan^-11 – tan^-1 0

= π/4

Concept:

\(\rm \int x^n dx=\frac{x^{n+1}}{n+1}+c\)

Calculation:

Let I = \(\rm \int_{0}^{\pi/4}\frac{\sin^{3}x}{\cos^{5}x}dx\)

\(\rm = \int_{0}^{\pi/4}\frac{\sin^{3}x}{\cos^{3}x}\times \frac{1}{\cos^2 x}dx\\=\int_{0}^{\pi/4} \tan^3 x\times \sec^2 x dx\)

Let tan x = t

Differentiating with respect to x, we get

⇒ sec² x dx =dt

Now,

\(\rm I = \int_{0}^{1} t^3 dt\\=\left[\frac{t^4}{4}\right]_0^1\\=\frac{1}{4}\)

Concept: 

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm\int {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C
• \(\rm \int{1\over 1+ x^2}\)dx = tan^-1 x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to is to be determined.
• Substitution is done and integral function is then integrated.
• Finally, initial variable t, to be returned.

Calculation:

Let sin x = t ⇒ cos x dx = dt

⇒ I = \(\rm \int {\cos x\over1+\sin^2x}\) dx

⇒ I = \(\rm \int{1\over 1+ t^2}\) dt

⇒ I = tan^-1 t + c

⇒ I = tan-1 (sin x) + c

Putting the limits

⇒ I = [tan-1 (sin 90º) + c] - [tan-1 (sin 0º) + c]

⇒ I = tan-1 (1)

⇒ I = 45º or π/4

Given:

\(\int_0^{\pi/4}\sin^3 \theta d\theta\)

Concept:

Use formula \(\rm sin^2\theta=1-cos^2\theta\)

Calculation:

\(\int_0^{\pi/4}\sin^3 \theta d\theta\)

\(\rm =\int_0^{\pi/4}(1-cos^2\theta)sin\theta \ d\theta\)

\(\rm =\int_0^{\pi/4}sin\theta \ d\theta-\int_0^{\pi/4}cos^2\theta\ sin\theta\ d \theta\)  

put \(\rm cos\theta=u \implies sin\theta=-d\theta\) 

\(\rm =\int_0^{\pi/4}sin\theta \ d\theta+\int_0^{\pi/4}u^2\ du\)

\(\rm =[-cos\theta]_0^{\pi/4}+[\frac{cos^{3} \theta}{3}]_0^{\pi/4}\)

\(\rm =-[\frac{1}{\sqrt2}-1]+\frac{1}{3}[\frac{1}{2\sqrt2}-1]\)

\(\rm =\frac{2}{3}-\frac{5}{6\sqrt2}\)

Hence the option (2) is correct.

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to be determined.
• Substitution is done and the integral function is then integrated.
• Finally, the initial variable t, to be returned.

Calculation:

I = \(\rm \int x\sin3x^2dx\)

Let 3x² = t; ⇒ 6x dx = dt

I = \(\rm {1\over6}\int \sin tdt\)

I = \(\rm -{1\over6} \cos t\)

I = \(\rm -{1\over6}\cos 3x^2\)

Putting up the limits
I = \(\rm \left[-{1\over6}\cos 3x^2\right]_0^{\pi\over6}\)

I = \(\rm -{1\over6}\left[\cos 3({\pi\over6})^2 - \cos 3(0)^2\right]\)

I = \(\rm -{1\over6}\left[\cos {\pi^2\over12} - 1\right]\)

I = \(\rm {1-\cos {\pi^2\over12}\over6}\)