Informative line

### Improper Integrals Type 2

Learn how to evaluate type 2 improper integral problems with comparison theorem for improper integrals, Practice Type 2 improper integral problems and infinite discontinuity calculus.

# Improper Integral

While evaluating $$\displaystyle \int\limits^b_a f(x)dx$$ there was an assumption that

(1) $$f$$ does not have an infinite discontinuity in $$[a, b]$$ .

(2) The integral $$[a, b]$$ is not infinite i.e, both $$a$$ and $$b$$ are finite values.

• If either of the above 2 condition is violated we say that the integral $$\displaystyle \int\limits^b_a f(x)dx$$  is an improper integral.

### Improper Integrals of Type 2

• If $$f$$ is continuous on $$[a,\,b)$$ and is discontinuous at $$x=b$$  then

$$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\lim\limits_{t\to b^-}\int\limits^t_af(x)\,dx$$

if this limit exists.

• If $$f$$ is continuous on $$(a,\,b]$$ and is discontinuous at $$x=a$$ then

$$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\lim\limits_{t\to a^+}\int\limits^b_tf(x)\,dx$$

If this limit exists.

The improper integral $$\displaystyle\int\limits^{b}_{a}f(x)\,dx$$ is called convergent if the corresponding limit exists and divergent if limit does not exist.

Shaded Area = $$\displaystyle\int\limits^{b}_{a}f(x)\,dx$$

If $$f(x)$$ has discontinuity at  $$x=c$$  when  $$a<c<b$$ then $$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\displaystyle\int\limits^{c}_{a}f(x)\,dx+\displaystyle\int\limits^{b}_{c}f(x)\,dx$$  if both of these converge.

#### Which of the following is an improper integral of type 2?

A $$I=\displaystyle \int\limits^7_4 \dfrac{1}{\sqrt{x-4}}dx$$

B $$I=\displaystyle \int\limits^5_{2} \dfrac{1}{\sqrt{x}}dx$$

C $$I=\displaystyle \int\limits^9_{4} \dfrac{1}{\sqrt{x+2}}dx$$

D $$I=\displaystyle \int\limits^3_1 \dfrac{1}{\sqrt{x+4}}dx$$

×

$$I=\displaystyle \int\limits^b_a f(x)dx$$ is an improper integral of type 2 if either $$f$$ is discontinuous at $$x=a$$ or at  $$x=b$$

Consider the options:

(a)  $$\to$$ $$\dfrac{1}{\sqrt{x-4}}$$ is discontinuous at $$x=4\to$$ improper

(b)  $$\dfrac{1}{\sqrt{x}}$$ is continuous in $$[2,\,5]\to$$ Proper

(c)  $$\dfrac{1}{\sqrt{x+2}}$$ is continuous in $$[4,\,9]\to$$ Proper

(d)  $$\dfrac{1}{\sqrt{x+4}}$$ is continuous in $$[1,\,3]\to$$ Proper

$$\therefore$$ Option A is correct

### Which of the following is an improper integral of type 2?

A

$$I=\displaystyle \int\limits^7_4 \dfrac{1}{\sqrt{x-4}}dx$$

.

B

$$I=\displaystyle \int\limits^5_{2} \dfrac{1}{\sqrt{x}}dx$$

C

$$I=\displaystyle \int\limits^9_{4} \dfrac{1}{\sqrt{x+2}}dx$$

D

$$I=\displaystyle \int\limits^3_1 \dfrac{1}{\sqrt{x+4}}dx$$

Option A is Correct

# Evaluation of Improper Integral Type 2

While evaluating $$\displaystyle \int\limits^b_a f(x)dx$$ there was an assumption that

(1) $$f$$ does not have an infinite discontinuity in $$[a, b]$$ .

(2) The integral $$[a, b]$$ is not infinite i.e, both $$a$$ and $$b$$ are finite values.

• If either of the above 2 condition is violated we say that the integral $$\displaystyle \int\limits^b_a f(x)dx$$ is an improper integral.

### Improper Integral of Type 2

• If $$f$$ is continuous on $$[a,\,b)$$ and is discontinuous at $$x=b$$ then

$$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\lim\limits_{t\to b^-}\int\limits^t_af(x)\,dx$$

If this limit exists.

• If $$f$$ is continuous on $$(a,\,b]$$ and is discontinuous at $$x=a$$ then

$$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\lim\limits_{t\to a^+}\int\limits^b_tf(x)\,dx$$

If this limit exists.

The improper integral $$\displaystyle\int\limits^{b}_{a}f(x)\,dx$$ is called convergent if the corresponding limit exists and divergent if limit does not exist.

Shaded Area= $$\int\limits_a^b f(x) \,dx$$

If $$f(x)$$ has discontinuity at $$x=c$$  when  $$a<c<b$$  then $$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\displaystyle\int\limits^{c}_{a}f(x)\,dx+\displaystyle\int\limits^{b}_{c}f(x)\,dx$$  if both of these converge.

#### Evaluate  $$I=\displaystyle \int\limits^5_2 \dfrac{3}{(x-2)^3}dx$$  if it converges.

A $$\dfrac{1}{2}$$

B $$\dfrac{-1}{3}$$

C 4

D It is divergent.

×

$$\displaystyle \int\limits^b_a f(x)dx=\lim\limits_{t\to a^+}\,\displaystyle \int\limits^b_t f(x)dx$$ if $$f$$ is discontinuous at $$x=a$$

$$\therefore\;\displaystyle \int\limits^5_2 \dfrac{3}{(x-2)^3}dx=\lim\limits_{t\to 2^+}\,\displaystyle \int\limits^5_t \dfrac{3}{(x-2)^3}dx$$

(The integrand is discontinuous at $$x=2$$)

$$\displaystyle \int\limits^5_t \dfrac{3}{(x-2)^3}dx=\Bigg[\dfrac{3(x-2)^{-2}}{-2}\Bigg]^5_t$$

$$=\dfrac{-3}{2}\left[(5-2)^2-(t-2)^2\right]$$

$$=\dfrac{-3}{2}\left[\dfrac{1}{9}-\dfrac{1}{(t-2)^2}\right]$$

$$=\dfrac{-3}{18}+\dfrac{3}{2(t-2)^2}$$

$$=\dfrac{-1}{6}+\dfrac{3}{2(t-2)^2}$$

$$\therefore\;\displaystyle \int\limits^5_2 \dfrac{3}{(x-2)^3}dx\\=\lim\limits_{t\to 2^+}\,\dfrac{-1}{6}+\dfrac{3}{(t-2)^2}$$

$$=\dfrac{-1}{6}+\infty=\infty$$

$$\therefore$$ doesn't converge

### Evaluate  $$I=\displaystyle \int\limits^5_2 \dfrac{3}{(x-2)^3}dx$$  if it converges.

A

$$\dfrac{1}{2}$$

.

B

$$\dfrac{-1}{3}$$

C

4

D

It is divergent.

Option D is Correct

# Finding the Value of a Parameter given that an Improper Integral of Type 1 Converges

While evaluating $$\displaystyle \int\limits^b_a f(x)dx$$ there was an assumption that

(1) $$f$$ does not have an infinite discontinuity in $$[a,\,b]$$ .

(2) The integral $$[a,\, b]$$ is not infinite i.e, both $$a$$ and $$b$$ are finite values.

• If either of the above 2 condition is violated we say that the integral $$\displaystyle \int\limits^b_a f(x)dx$$ is an improper integral.

## Improper Integrals of Type 1

• If $$\displaystyle \int\limits^t_a f(x)dx$$ exist for every real number $$t \geq a$$ then $$\displaystyle \int\limits^\infty_a f(x)dx= \lim_{t \to \infty} \displaystyle \int\limits^t_a f(x)dx$$ provided this limit exists.
• If $$\displaystyle \int\limits^b_t f(x)dx$$ exist for every real number $$t \leq b$$ then $$\displaystyle \int\limits^b_{-\infty} f(x)dx= \lim_{t \to -\infty} \displaystyle \int\limits^b_t f(x)dx$$ provided this limit exists.
• These improper integrals $$\displaystyle \int\limits^\infty_a f(x)dx$$  and  $$\displaystyle \int\limits^b_{-\infty} f(x)dx$$ are called convergent if corresponding limit exists.
• By existence of limit we mean that

$$\displaystyle \int\limits^\infty_a f(x)dx= \lim_{t \to \infty} \displaystyle \int\limits^t_a f(x)dx$$

is a finite value.

and $$\displaystyle \int\limits^b_{-\infty} f(x)dx= \lim_{t \to -\infty} \displaystyle \int\limits^b_t f(x)dx$$

is a finite value

and  divergent if corresponding limit does not exist.

#### For what value of $$d$$ will integral  $$I=\displaystyle \int\limits^\infty_0 \left(\dfrac{2x}{x^2+1}-\dfrac{d}{2x+1}\right)dx$$ converge?

A $$d=2$$

B $$d=4$$

C $$d=1$$

D $$d=6$$

×

$$\displaystyle \int\limits^\infty_a f(x)dx$$  converges if  $$\lim\limits_{t\to \infty}\displaystyle \int\limits^b_a f(x)dx$$ exists

$$I=\displaystyle \int\limits^\infty_0 \left(\dfrac{2x}{x^2+1}-\dfrac{d}{2x+1}\right)dx$$

$$=\lim\limits_{t\to\infty}\displaystyle \int\limits^t_0 \left(\dfrac{2x}{x^2+1}-\dfrac{d}{2x+1}\right)dx$$

Now consider,

$$\displaystyle\int\, \Bigg[\underbrace{\dfrac{2x}{x^2+1}}_ {put\;x^2+1=t}-\dfrac{d}{2x+1}\Bigg]dx$$

$$=ln(x^2+1)-\dfrac{d\,ln(2x+1)}{2}\\=ln\underbrace{\dfrac{x^2+1}{(2x+1)^{d/2}}}_\text{Properties of log}$$

$$\displaystyle \int\limits^t_0 \left(\dfrac{2x}{x^2+1}-\dfrac{d}{2x+1}\right)dx$$

$$=ln\dfrac{t^2+1}{(2t+1)^{d/2}}-ln\,1$$

$$=ln\dfrac{t^2+1}{(2t+1)^{d/2}}$$

$$\therefore\;I=\lim\limits_{t\to\infty}ln\dfrac{t^2+1}{(2t+1)^{d/2}}$$ should be finite this will happen when the domain $$(2t+1)^{d/2}$$ is a quadratic in $$t$$ i.e. $$d=4$$.

when $$d=4$$,

$$I=\lim\limits_{t\to\infty}ln\dfrac{t^2+1}{(2t+1)^{2}}$$

$$=\lim\limits_{t\to\infty}ln\dfrac{t^2+1}{4t^2+4t+1}$$

$$=ln\,\dfrac{1}{4}$$

### For what value of $$d$$ will integral  $$I=\displaystyle \int\limits^\infty_0 \left(\dfrac{2x}{x^2+1}-\dfrac{d}{2x+1}\right)dx$$ converge?

A

$$d=2$$

.

B

$$d=4$$

C

$$d=1$$

D

$$d=6$$

Option B is Correct

# Important integral

Important integral $$I_n=\displaystyle\int\limits^\infty_0x^ne^{-x}dx$$,  where $$n$$ is a positive integral

Consider,

$$I_n=\displaystyle\int\limits^\infty_0x^ne^{-x}dx$$

$$=\Bigg[-x^ne^{-x}\Bigg]^\infty _0+\displaystyle\int\limits^\infty_0nx^{n-1}e^{-x}dx$$                  (Integration by parts)

$$=\Bigg[-x^ne^{-x}\Bigg]^\infty_0+n\displaystyle\int\limits^\infty_0x^{n-1}e^{-x}dx$$

$$=\Bigg[-x^ne^{-x}\Bigg]^\infty_0+nI_{n-1}$$

$$\therefore\;I_n=\Bigg[-x^ne^{-x}\Bigg]^\infty_0+nI_{n-1}$$

$$=(0-0)+ n\displaystyle\int\limits^\infty_0x^{n-1}e^{-x}dx$$       $$\left(\lim\limits_{t\to\infty}-\dfrac{t^n}{e^t}=0\right)$$

$$\therefore\;I_n=nI_{n-1}$$

e.g.  $$\displaystyle\int\limits^\infty_0x^{2}e^{-x}dx=I_2=2I_1=2×1×I_0$$

and $$I_0=\displaystyle\int\limits^\infty_0e^{-x}dx\\=\Bigg[-e^{-x}\Bigg]^\infty_0\\=[0-1]=1$$

$$\therefore\;I_0=1$$

$$\therefore\;I_2=2I_0=2$$

#### Evaluate  $$I=\displaystyle\int\limits^\infty_0x^{4}e^{-x}dx$$

A 15

B 24

C 9

D 6

×

If $$I_n=\displaystyle\int\limits^\infty_0x^{n}e^{-x}dx$$

then $$I_n=nI_{n-1}$$

$$\therefore\;I_4=4×I_3\\=4×3×I_2\\=4×3×2×I_1$$

$$=4×3×2×1×I_0$$

$$=4×3×2×1×1=24$$

(Put $$x=4,\;3,\;2$$ and $$1$$ in the above formula)

$$\therefore\;I_4=24$$

### Evaluate  $$I=\displaystyle\int\limits^\infty_0x^{4}e^{-x}dx$$

A

15

.

B

24

C

9

D

6

Option B is Correct

# Convergent and Divergent Integrals

• Improper integral $$\int\limits_a^{\infty}\,f(x)\;dx$$ is said to be convergent if $$\lim\limits_{t\rightarrow \infty}\int\limits_a^t\,f(x)\;dx$$ exists or is a finite value
• If it is not convergent then we can say it is divergent.
• Improper integral $$\int\limits_{-\infty}^b\,f(x)\;dx$$ is convergent if $$\lim\limits_{t\rightarrow -\infty}\int\limits_t^b\,f(x)\;dx$$ exists or is a finite value.
• If it is not convergent we say that it is divergent.

#### Which of the following is correct about the following integrals? $$I_1=\int\limits_2^{\infty}\,\dfrac {1}{x^2}\;dx$$ and  $$I_2=\int\limits_2^{\infty}\,\dfrac {1}{\sqrt x}\;dx$$

A $$I_1$$ converges but $$I_2$$ diverges

B $$I_1$$ diverges but $$I_2$$ converges

C $$I_1$$ and $$I_2$$ both are converges

D $$I_1$$ and $$I_2$$ both are diverges.

×

$$\int\limits_a^{\infty}\,f(x)\;dx=\lim\limits_{t\rightarrow \infty}\int\limits_a^t\,f(x)\;dx$$ converges if the limits exists.

$$\int\limits_{-\infty}^b\,f(x)\;dx=\lim\limits_{t\rightarrow -\infty}\int\limits_t^b\,f(x)\;dx$$ converges if the limits exists.

In this case,

$$I_1=\int\limits_2^{\infty}\,\dfrac {1}{x^2}\;dx=\lim\limits_{t\rightarrow \infty}\int\limits_2^t\,\dfrac {1}{x^2}\;dx$$

$$=\lim\limits_{t\rightarrow \infty}\;\Big[\dfrac {-1}{x}\Big]_2^t\\= \lim\limits_{t\rightarrow \infty}\;\Big[\dfrac {-1}{t}+\dfrac {1}{2}\Big]\\=\dfrac {1}{2}$$

$$\therefore \;I_1$$ converges

$$I_2=\int\limits_2^{\infty}\,\dfrac {1}{\sqrt x}\;dx \\=\int\limits_2^{\infty}\,x^{-1/2}\;dx\\=\;\Big[\dfrac {x^{1/2}}{1/2}\Big]_2^{\infty}$$

$$=\lim\limits_{t\rightarrow \infty}\;\Big[2\sqrt x\Big]_2^t\\=2 \lim\limits_{t\rightarrow \infty}\;\Big[\sqrt t -\sqrt 2\Big]\\=\infty$$

$$\therefore \;I_2$$ diverges

$$\therefore$$ Correct option is (A).

### Which of the following is correct about the following integrals? $$I_1=\int\limits_2^{\infty}\,\dfrac {1}{x^2}\;dx$$ and  $$I_2=\int\limits_2^{\infty}\,\dfrac {1}{\sqrt x}\;dx$$

A

$$I_1$$ converges but $$I_2$$ diverges

.

B

$$I_1$$ diverges but $$I_2$$ converges

C

$$I_1$$ and $$I_2$$ both are converges

D

$$I_1$$ and $$I_2$$ both are diverges.

Option A is Correct

# Evaluation of Improper Integral (Type 2)

While evaluating $$\displaystyle \int\limits^b_a f(x)dx$$ there was an assumption that

(1) $$f$$ does not have an infinite discontinuity in $$[a, b]$$ .

(2) The integral $$[a, b]$$ is not infinite i.e, both $$a$$ and $$b$$ are finite values.

• If either of the above 2 condition is violated we say that the integral $$\displaystyle \int\limits^b_a f(x)dx$$ is an improper integral.

### Improper Integral of type 2

• If $$f$$ is continuous on $$[a,\,b)$$ and is discontinuous at $$x=b$$ then

$$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\lim\limits_{t\to b^-}\int\limits^t_af(x)\,dx$$

If this limit exists.

• If $$f$$ is continuous on $$(a,\,b]$$ and is discontinuous at $$x=a$$ then

$$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\lim\limits_{t\to a^+}\int\limits^b_t\,f(x)\,dx$$

If this limit exists.

The improper integral $$\displaystyle\int\limits^{b}_{a}f(x)\,dx$$ is called convergent if the corresponding limit exists and divergent if limit does not exist.

Shaded Area $$=\int\limits_a^b\, f(x)\,dx$$

If $$f(x)$$ has discontinuity at $$x=c$$ when $$a<c<b$$ then $$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\displaystyle\int\limits^{c}_{a}f(x)\,dx+\displaystyle\int\limits^{b}_{c}f(x)\,dx$$ if both of these convergent.

#### Evaluate  $$I=\displaystyle \int\limits^2_1 \dfrac{x}{\sqrt{x-1}}dx$$ if it converges.

A $$\dfrac{5}{4}$$

B $$\dfrac{8}{3}$$

C $$\dfrac{-1}{2}$$

D 2

×

$$\displaystyle \int\limits^b_a f(x)dx=\lim\limits_{t\to a^+}\,\displaystyle \int\limits^b_t f(x)dx$$ if $$f$$ is discontinuous at $$x=a$$

$$\dfrac{x}{\sqrt{x-1}}dx$$ is discontinuous at $$x=1$$

$$\therefore\;\displaystyle \int\limits^2_1 \dfrac{x}{\sqrt{x-1}}dx=\lim\limits_{t\to 1^+}\,\displaystyle \int\limits^2_t \dfrac{x}{\sqrt{x-1}}dx$$

Now consider,

$$I_1=\displaystyle \int\limits\dfrac{x}{\sqrt{x-1}}dx\,\,\,\,\,\,\to$$ put $$x-1=x^2$$

$$\Rightarrow\;dx=2x\,dx$$

$$\therefore\;I_1=\displaystyle \int\limits \dfrac{(x^2+1)×2x\,dx}{x}$$

$$=\displaystyle \int\limits(x^2+1)×2\,dx$$

$$=2\left(\dfrac{x^3}{3}+x\right)$$

$$=2\left(\dfrac{(x-1)^{3/2}}{3}\right)+2\;(x-1)^{1/2}$$

$$\therefore\;\displaystyle \int\limits^2_t \dfrac{x}{\sqrt{x-1}}dx$$

$$=2\left[\dfrac{(x-1)^{3/2}}{3}+(x-1)^{1/2}\right]^2_t$$

$$=2\left[\left(\dfrac{1}{3}+1\right)-\left(\dfrac{(t-1)^{3/2}}{3}+(t-1)^{1/2}\right)\right]$$

$$=\dfrac{8}{3}-\dfrac{2}{3}(t-1)^{3/2}-2(t-1)^{1/2}$$

$$\therefore\;\displaystyle\int\limits^2_1\dfrac{x}{\sqrt{x-1}}dx=\lim\limits_{t\to 1^+}\left(\dfrac{8}{3}-\underbrace{\dfrac{2}{3}(t-1)^{3/2}}_{0\;as\;t\to1^+}-\underbrace{2(t-1)^{1/2}}_{0\;as\;t\to1^+}\right)$$

$$=\dfrac{8}{3}-0-0$$

$$=\dfrac{8}{3}$$

### Evaluate  $$I=\displaystyle \int\limits^2_1 \dfrac{x}{\sqrt{x-1}}dx$$ if it converges.

A

$$\dfrac{5}{4}$$

.

B

$$\dfrac{8}{3}$$

C

$$\dfrac{-1}{2}$$

D

2

Option B is Correct

# Improper Integrals

While evaluating $$\displaystyle \int\limits^b_a f(x)dx$$ there was an assumption that

(1) $$f$$ does not have an infinite discontinuity in $$[a, b]$$ .

(2) The integral $$[a, b]$$ is not infinite i.e, both $$a$$ and $$b$$ are finite values.

• If either of the above 2 condition is violated we say that the integral $$\displaystyle \int\limits^b_a f(x)dx$$ is an improper integral.

### Improper Integral of type 2

• If $$f$$ is continuous on $$[a,\,b)$$ and is discontinuous at $$x=b$$ then

$$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\lim\limits_{t\to b^-}\int\limits^t_af(x)\,dx$$

If this limit exists.

• If $$f$$ is continuous on $$(a,\,b]$$ and is discontinuous at $$x=a$$ then

$$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\lim\limits_{t\to a^+}\int\limits^b_t \,f(x)\,dx$$

If this limit exists.

The improper integral $$\displaystyle\int\limits^{b}_{a}f(x)\,dx$$ is called convergent if the corresponding limit exists and divergent if limit does not exist.

Shaded Area= $$\displaystyle\int\limits^{b}_{a}f(x)\,dx$$

If $$f(x)$$ has discontinuity at $$x=c$$  when $$a<c<b$$  then $$\displaystyle\int\limits^{b}_{a}f(x)\,dx=\displaystyle\int\limits^{c}_{a}f(x)\,dx+\displaystyle\int\limits^{b}_{c}f(x)\,dx$$  if both of these converge.

#### Evaluate  $$I=\displaystyle \int\limits^1_{-1} \dfrac{x+1}{\sqrt[5]{x^3}}dx$$ if it converges.

A $$\dfrac{5}{7}$$

B $$\dfrac{10}{7}$$

C $$\dfrac{3}{4}$$

D $$\dfrac{1}2{}$$

×

If $$f(x)$$ has a discontinuity at $$x=c$$ where $$a<c<b$$ then $$\displaystyle \int\limits^b_a f(x)dx=\displaystyle \int\limits^c_a f(x)dx+\displaystyle \int\limits^b_c f(x)dx$$ if both converge

In this case $$f(x)=\dfrac{x+1}{x^{3/5}}$$ is discontinuous at $$x=0$$

$$\therefore\;I=\displaystyle \int\limits^0_{-1} \underbrace{\dfrac{x+1}{x^{3/5}}}_{I_1}dx=\displaystyle \int\limits^1_0 \underbrace{\dfrac{x+1}{x^{3/5}}}_{I_2}dx$$

$$I_1=\displaystyle \int\limits^0_{-1}\dfrac{x+1}{x^{3/5}}dx=\lim\limits_{t\to0^-}\,\displaystyle \int\limits^t_{-1}\dfrac{x+1}{x^{3/5}}dx$$

$$=\lim\limits_{t\to0^-}\,\displaystyle \int\limits^t_{-1}\left(x^{2/5}+x^{-3/5}\right)dx=\lim\limits_{t\to0^-}\dfrac{x^{7/5}}{\dfrac{7}{5}}+\dfrac{x^{2/5}}{\dfrac{2}{5}}\Bigg]^t_{-1}$$

$$=\lim\limits_{t\to0^-}\,\left(\dfrac{5}{7}\,t^{7/5}+\dfrac{5}{2}\,t^{2/5}\right)-\left(-\dfrac{5}{7}+\dfrac{5}{2}\right)$$

$$=\lim\limits_{t\to0^-}\,\left(\dfrac{5}{7}\,t^{7/5}+\dfrac{5}{2}\,t^{2/5}-\dfrac{25}{14}\right)=-\dfrac{25}{14}$$

$$I_2=\displaystyle \int\limits^1_{0}\dfrac{x+1}{x^{3/5}}dx\\=\lim\limits_{t\to0^+}\,\displaystyle \int\limits^1_{t}\left(x^{2/5}+x^{-3/5}\right)dx$$

$$=\lim\limits_{t\to0^+}\dfrac{x^{7/5}}{\dfrac{7}{5}}+\dfrac{x^{2/5}}{\dfrac{2}{5}}\Bigg]^1_{t}$$

$$=\lim\limits_{t\to0^+}\,\left(\dfrac{5}{7}+\dfrac{5}{2}\right)-\underbrace{\left(\dfrac{5}{7}\,t^{7/5}+\dfrac{5}{2}\,t^{2/5}\right)}_\text{equals to 0}$$

$$=\dfrac{45}{14}-0-0$$

$$=\dfrac{45}{14}$$

$$\therefore\;I=I_1+I_2\\=-\dfrac{25}{14}+\dfrac{45}{14}$$

$$=\dfrac{20}{14}\\=\dfrac{10}{7}$$

### Evaluate  $$I=\displaystyle \int\limits^1_{-1} \dfrac{x+1}{\sqrt[5]{x^3}}dx$$ if it converges.

A

$$\dfrac{5}{7}$$

.

B

$$\dfrac{10}{7}$$

C

$$\dfrac{3}{4}$$

D

$$\dfrac{1}2{}$$

Option B is Correct

# Comparison Theorem for Improper Integrals

Sometimes we are not able to find the value of an improper integral, but we are interested in knowing whether it is divergent or convergent.

• For such improper integrals we use the comparison theorem.

### Comparison Theorem :

Suppose $$f$$ and $$g$$ are continuous function with $$f(x)\geq g(x)\geq0$$ for $$x\geq a$$ then,

If area under the higher curve $$f(x)$$ is finite then so is the area under the lower curve.

(1)  If $$\displaystyle\int\limits^\infty_a f(x)dx$$ is convergent then $$\displaystyle\int\limits^\infty_a g(x)dx$$ is also convergent.

(2)  If $$\displaystyle\int\limits^\infty_a g(x)dx$$ is divergent then $$\displaystyle\int\limits^\infty_a f(x)dx$$ is divergent.

#### Consider, Two improper integrals $$I_1=\displaystyle\int\limits^\infty_1\,\dfrac{5+e^{-x^2}}{x}dx$$ and $$I_2=\displaystyle\int\limits^1_0\,\dfrac{sec^2x}{x^2\sqrt x}$$ choose the correct option for $$I_1\;\&\;I_2$$.

A $$I_1$$ and $$I_2$$ are both convergent.

B $$I_1$$ and $$I_2$$ are both divergent.

C $$I_1$$ is convergent but $$I_2$$ is divergent.

D $$I_1$$ is divergent but $$I_2$$ is convergent

×

If $$f(x)\geq g(x)$$ for $$x\geq a$$ then

(1)  $$\displaystyle\int\limits^\infty_a f(x)dx$$ is convergent

$$\Rightarrow\;\displaystyle\int\limits^\infty_a g(x)dx$$ is convergent

(2)  $$\displaystyle\int\limits^\infty_a g(x)dx$$ is divergent

$$\Rightarrow\;\displaystyle\int\limits^\infty_a f(x)dx$$ is divergent.

Consider  $$I_1\to f(x)=\dfrac{5+e^{-x^2}}{x}=\dfrac{5}{x}+\dfrac{e^{-x^2}}{x}>\dfrac{5}{x}$$ in $$(1,\;\infty)$$

Now $$\displaystyle\int\limits^\infty_1\,\dfrac{5}{x}dx=5ln\,x\Bigg|^\infty_1=\infty\Rightarrow$$ divergent

$$\therefore\;I_1$$ is divergent (By comparison theorem)

Consider ,

$$I_2\to \dfrac{sec^2x}{x^2\sqrt x}=\dfrac{1+tan^2x}{x^2\sqrt x}$$

$$=\dfrac{1}{x^2\sqrt x}+\dfrac{tan^2x}{x\sqrt x}>\dfrac{1}{x^2\sqrt x}$$ in $$(0,\;1)$$

Now $$\displaystyle\int\limits^1_0 \dfrac{1}{x^2\sqrt x}dx\\=\displaystyle\int\limits^1_0 x^{-5/2}dx\\=\Bigg[\dfrac{x^{-3/2}}{\dfrac{-3}{2}}\Bigg]^1_0\\=-\dfrac{2}{3}[1-\infty]$$

$$=\infty\Rightarrow\;I_2$$ is also divergent (By comparison theorem)

$$\therefore$$ correct option is  $$(B)$$.

### Consider, Two improper integrals $$I_1=\displaystyle\int\limits^\infty_1\,\dfrac{5+e^{-x^2}}{x}dx$$ and $$I_2=\displaystyle\int\limits^1_0\,\dfrac{sec^2x}{x^2\sqrt x}$$ choose the correct option for $$I_1\;\&\;I_2$$.

A

$$I_1$$ and $$I_2$$ are both convergent.

.

B

$$I_1$$ and $$I_2$$ are both divergent.

C

$$I_1$$ is convergent but $$I_2$$ is divergent.

D

$$I_1$$ is divergent but $$I_2$$ is convergent

Option B is Correct