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### Definition Of Improper Integral And Type 1

Learn how to evaluating improper integrals problems with Infinite Discontinuity, Practice Type 1 improper integral problems, infinite discontinuity calculus functions.

# 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 Integrals of Type 1

• If $$\displaystyle \int\limits^t_a f(x)dx$$ exists 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 $$\displaystyle \int\limits^b_t f(x)dx$$ exists 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 exists.

#### Which of the following integral is improper ?

A $$I=\displaystyle \int\limits^4_2 (x+3)dx$$

B $$I=\displaystyle \int\limits^5_{-2} (x^2+5)dx$$

C $$I=\displaystyle \int\limits^7_{-5} (x^2+5)dx$$

D $$I=\displaystyle \int\limits^\infty_2 \dfrac{1}{x^2}dx$$

×

Integrals of the form  $$\displaystyle \int\limits^\infty_a f(x)dx$$  and  $$\displaystyle \int\limits^b_{-\infty} f(x)dx$$ are called improper.

$$\therefore$$ Option (D) is an improper integral.

### Which of the following integral is improper ?

A

$$I=\displaystyle \int\limits^4_2 (x+3)dx$$

.

B

$$I=\displaystyle \int\limits^5_{-2} (x^2+5)dx$$

C

$$I=\displaystyle \int\limits^7_{-5} (x^2+5)dx$$

D

$$I=\displaystyle \int\limits^\infty_2 \dfrac{1}{x^2}dx$$

Option D 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 1

• If $$\displaystyle \int\limits^t_a f(x)dx$$ exists 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 $$\displaystyle \int\limits^b_t f(x)dx$$ exists 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 integral $$\displaystyle \int\limits^\infty_a f(x)dx$$  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 exists.

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

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

B $$I=\displaystyle \int\limits^{-2}_{-4} \dfrac{1}{x^2}dx$$

C $$I=\displaystyle \int\limits^2_{–\,\infty} \dfrac{1}{x^2}dx$$

D $$I=\displaystyle \int\limits^{-1}_{-5} \dfrac{1}{x^3}dx$$

×

Integrals of the form  $$\displaystyle \int\limits^\infty_a f(x)dx$$  and  $$\displaystyle \int\limits^b_{-\infty} f(x)dx$$ are called improper.

$$\therefore$$Option (C) is correct.

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

A

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

.

B

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

C

$$I=\displaystyle \int\limits^2_{–\,\infty} \dfrac{1}{x^2}dx$$

D

$$I=\displaystyle \int\limits^{-1}_{-5} \dfrac{1}{x^3}dx$$

Option C 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 1

• If $$\displaystyle \int\limits^t_a f(x)dx$$ exists 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 $$\displaystyle \int\limits^b_t f(x)dx$$ exists 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 integral $$\displaystyle \int\limits^\infty_a f(x)dx$$  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 exists.

#### Evaluate  $$\text I=\displaystyle \int\limits^\infty_1 \dfrac{dx}{x^4}$$  if it converges.

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

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

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

D $$\dfrac{1}{4}$$

×

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

$$\therefore \displaystyle \int\limits^\infty_1 \dfrac{1}{x^4}dx= \lim_{t \to \infty} \displaystyle \int\limits^t_1 \dfrac{1}{x^4}dx$$

Now

$$\therefore \displaystyle \int\limits^t_1 \dfrac{1}{x^4}dx\\= \displaystyle \int\limits^t_1 x^{-4}\;dx\\= \left[\dfrac{x^{-3}}{-3}\right]_1^t$$

$$=-\dfrac{1}{3}\left[\dfrac{1}{t^3}-1\right] \\=\dfrac{1}{3}\left[1-\dfrac{1}{t^3}\right]$$

$$\therefore \;\displaystyle \int\limits^\infty_1 \dfrac{1}{x^4}dx\\=\lim\limits_{t \to \infty}\dfrac{1}{3}\left(1-\dfrac{1}{t^3}\right)\\=\dfrac{1}{3}(1-0)\\=\dfrac{1}{3}$$

### Evaluate  $$\text I=\displaystyle \int\limits^\infty_1 \dfrac{dx}{x^4}$$  if it converges.

A

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

.

B

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

C

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

D

$$\dfrac{1}{4}$$

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 Integrals of Type 1

• If $$\displaystyle \int\limits^t_a f(x)dx$$ exists 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 $$\displaystyle \int\limits^b_t f(x)dx$$ exists 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 exists.

#### Evaluate $$I=\displaystyle \int\limits^\infty_1 \dfrac{dx}{x^2+2x+5}$$  if it converges.

A $$\dfrac{\pi}{4}$$

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

C $$\dfrac{\pi}{6}$$

D $$\dfrac{\pi}{8}$$

×

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

$$\therefore \displaystyle \int\limits^\infty_1 \dfrac{dx}{x^2+2x+5}\\= \lim\limits _{t \to \infty} \displaystyle \int\limits^t_1 \dfrac{dx}{x^2+2x+5}$$

Now

$$\displaystyle \int \dfrac{dx}{x^2+2x+5}\\= \underbrace{\displaystyle \int \dfrac{dx}{(x+1)^2+4}}_\text{complete the square}$$

$$=\dfrac{1}{2}tan^{-1}\dfrac{x+1}{2}$$      (Use $$\displaystyle\int\dfrac{1}{x^2+a^2}dx=\dfrac{1}{a}tan^{-1}\dfrac{x}{a}$$ )

$$\therefore \displaystyle \int\limits^t_1 \dfrac{dx}{x^2+2x+5}\\= \dfrac{1}{2}\left[tan^{-1}\dfrac{t+1}{2}-tan^{-1}1\right]$$

$$\therefore \displaystyle \int\limits^\infty_1 \dfrac{dx}{x^2+2x+5} \\= \lim\limits _{t \to \infty}\dfrac{1}{2}\left[tan^{-1}\dfrac{t+1}{2}-\dfrac{\pi}{4}\right]$$

$$= \dfrac{1}{2}\left[\dfrac{\pi}{2}-\dfrac{\pi}{4}\right]\\=\dfrac{\pi}{8}$$

### Evaluate $$I=\displaystyle \int\limits^\infty_1 \dfrac{dx}{x^2+2x+5}$$  if it converges.

A

$$\dfrac{\pi}{4}$$

.

B

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

C

$$\dfrac{\pi}{6}$$

D

$$\dfrac{\pi}{8}$$

Option D 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 Integrals of Type 1

• If $$\displaystyle \int\limits^t_a f(x)dx$$ exists 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 $$\displaystyle \int\limits^b_t f(x)dx$$ exists 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 exists.

#### Evaluate $$I=\displaystyle \int\limits^0_{-\infty} x\;e^{3x}\;dx$$ if it converges.

A $$-\dfrac{1}{9}$$

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

C $$\dfrac{1}{3}$$

D $$-\dfrac{1}{6}$$

×

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

$$\therefore\displaystyle \int\limits^0_{-\infty} xe^{3x}dx= \lim_{t \to\, -\infty} \displaystyle \int\limits^0_t xe^{3x}dx$$

Now

$$\displaystyle \int xe^{3x}dx= \dfrac{xe^{3x}}{3} - \displaystyle \int\dfrac{e^{3x}}{3}dx$$

$$= \dfrac{xe^{3x}}{3}-\dfrac{e^{3x}}{9}\,\,\,\,\rightarrow$$ ( By Integration by parts)

$$\therefore \displaystyle \int\limits^0_t xe^{3x}dx=\left[ \dfrac{xe^{3x}}{3}-\dfrac{e^{3x}}{9}\right]_t^0\\=\left(-\dfrac{1}{9}\right)-\left( \dfrac{te^{3t}}{3}-\dfrac{e^{3t}}{9}\right)$$

$$\therefore \displaystyle \int\limits^0_{-\infty} xe^{3x}dx=\lim_{t \to\,- \infty}\left(-\dfrac{1}{9}-\dfrac{te^{3t\nearrow^{0\;as\;t\to-\infty}}}{3}+\dfrac{e^{3t}}{9}\right)$$

$$=-\dfrac{1}{9}-0+0\\=-\dfrac{1}{9}$$

### Evaluate $$I=\displaystyle \int\limits^0_{-\infty} x\;e^{3x}\;dx$$ if it converges.

A

$$-\dfrac{1}{9}$$

.

B

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

C

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

D

$$-\dfrac{1}{6}$$

Option A 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 Integrals of Type 1

• If $$\displaystyle \int\limits^t_a f(x)dx$$ exists 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 $$\displaystyle \int\limits^b_t f(x)dx$$ exists 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 exists.

If both  $$\displaystyle \int\limits^a_{-\infty} f(x)dx$$ and $$\displaystyle \int\limits^\infty_a f(x)dx$$  are convergent then

$$\displaystyle \int\limits^\infty_{-\infty} f(x)dx=\displaystyle \int\limits^a_{-\infty} f(x)dx+\displaystyle \int\limits^\infty_a f(x)dx$$

where 'a' can be any real number.

#### Evaluate $$I = \displaystyle \int\limits^\infty_{-\infty} \dfrac{x^4}{x^{10}+1}dx$$  if it converges.

A $$\dfrac{\pi}{10}$$

B $$\dfrac{\pi}{5}$$

C $$\dfrac{\pi^2}{3}$$

D $$\pi$$

×

$$\displaystyle \int\limits^\infty_{-\infty} f(x)dx=\displaystyle \int\limits^a_{-\infty} f(x)dx+\displaystyle \int\limits^\infty_a f(x)dx$$

where 'a' is any real number.

$$\displaystyle \int\limits^\infty_{-\infty} \dfrac{x^4}{x^{10}+1}dx =\displaystyle \int\limits^0_{-\infty} \dfrac{x^4}{x^{10}+1}dx +\displaystyle \int\limits^\infty_0 \dfrac{x^4}{x^{10}+1} dx$$    $$(\text {change} \,\,a=0)$$

Now consider,

$$I=\displaystyle \int \dfrac{x^4}{x^{10}+1}dx$$

Put $$x^5=U$$ and

$$5x^4dx=dU$$

We made this substitution as we know that $$\dfrac{d}{dx}x^5=5x^4$$ and it is present in the numerator.

$$\implies I=\dfrac{1}{5}\displaystyle\int\dfrac{dU}{U^2+1}\\=\dfrac{1}{5}tan^{-1}\;U\\=\dfrac{1}{5}tan^{-1}\;x^5$$

$$\therefore \; \displaystyle \int\limits^0_{-\infty} \dfrac{x^4}{x^{10}+1}dx =\lim_{t \to {-\infty}}\displaystyle \int\limits^0_t \dfrac{x^4}{x^{10}+1}dx$$

$$\displaystyle\lim_{t \to {-\infty}} \; \left[\dfrac{1}{5}tan^{-1}\;x^5\right]_t^0\\=\displaystyle\lim_{t \to {-\infty}} \left(0-\dfrac{1}{5}tan^{-1}\;t^5\right)$$

$$=-\dfrac{1}{5}×tan^{-1}(-\infty)\\=\dfrac{1}{5}×\dfrac{\pi}{2}=\dfrac{\pi}{10}$$

$$\therefore \; \displaystyle \int\limits^\infty_0 \dfrac{x^4}{x^{10}+1}dx\\ =\lim\limits_{t \to \infty}\displaystyle \int\limits^t_0 \dfrac{x^4}{x^{10}+1}dx$$

$$=\displaystyle\lim_{t \to \infty} \; \left[\dfrac{1}{5}tan^{-1}\;x^5\right]_0^t\\=\displaystyle\lim_{t \to \infty} \left(\dfrac{1}{5}tan^{-1}\;t^5\right)$$

$$\dfrac{1}{5}×\dfrac{\pi}{2}=\dfrac{\pi}{10}$$

$$\displaystyle \int\limits^\infty_{-\infty} \dfrac{x^4}{x^{10}+1}dx\\ =\dfrac{\pi}{10}+\dfrac{\pi}{10}\\=\dfrac{2\pi}{10}\\=\dfrac{\pi}{5}$$

### Evaluate $$I = \displaystyle \int\limits^\infty_{-\infty} \dfrac{x^4}{x^{10}+1}dx$$  if it converges.

A

$$\dfrac{\pi}{10}$$

.

B

$$\dfrac{\pi}{5}$$

C

$$\dfrac{\pi^2}{3}$$

D

$$\pi$$

Option B is Correct

# Improper Integrals of Type 1 as an Area

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 $$\displaystyle \int\limits^t_a f(x)dx$$ exists 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 $$\displaystyle \int\limits^b_t f(x)dx$$ exists 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 exists.

For a positive function $$f$$ we know that

$$\displaystyle \int\limits^b_a f(x)dx$$ = Area bounded by the curve

$$y=f(x)$$, lines $$x=a,\;x=b$$ and $$x$$ axis.

If $$\displaystyle \int\limits^\infty_a f(x)dx$$ is convergent we say that it is the area of region such that $$x \geq a$$  and $$0 \leq y \leq f(x)$$

$$A(S)=\displaystyle \int\limits^\infty_a f(x)dx$$

Note that the area bounded by  $$y=f(x)$$$$x$$  axis,

$$x=a$$  and  $$x=b$$  is given by

$$\displaystyle \int\limits^b_a f(x)dx$$

In this case  $$b \to \infty$$.

#### Sketch the region $$S={\{(x,y)}|x \geq1,\;0 \leq y \leq e^{-2x}\}$$  and find its area.

A $$A(S)=2$$

B $$A(S)=10$$

C $$A(S)=\dfrac{1}{2}$$

D $$A(S)=1$$

×

$$y=e^{-2x}$$  is exponential decay curve.

$$\therefore 0 \leq y \leq e^{-2x}$$ is the shaded region.

Now $$x \geq1$$ and $$0 \leq y \leq e^{-2x}$$ will be the region given below.

Shaded Area = $$A(S)=\displaystyle \int\limits^\infty_1 e^{-2x}\;dx$$

$$=\displaystyle\lim_{t \to \infty} \int\limits^t_1 e^{-2x}\;dx\\=\displaystyle\lim_{t \to \infty} \left[\dfrac{e^{-2x}}{-2} \right]_0^t$$

$$=\displaystyle\lim_{t \to \infty} -\dfrac{1}{2} [e^{-2t}-1]\\=\lim\limits_{t \to \infty} \left(\dfrac{1}{2}-\dfrac{e^{-2t\nearrow^{0 \;as\;t\to\infty}}}{-2} \right)$$

$$=\dfrac{1}{2}-0\\=\dfrac{1}{2}$$

### Sketch the region $$S={\{(x,y)}|x \geq1,\;0 \leq y \leq e^{-2x}\}$$  and find its area.

A

$$A(S)=2$$

.

B

$$A(S)=10$$

C

$$A(S)=\dfrac{1}{2}$$

D

$$A(S)=1$$

Option C is Correct

# Area as an Improper Integral

For a positive function $$f$$ we know that,

$$\displaystyle \int\limits^b_a f(x)dx$$ = Area bounded by the curve

$$y=f(x)$$, lines $$x=a,\;x=b$$ and $$x$$ axis.

• If $$\displaystyle \int\limits^\infty_a f(x)dx$$ is convergent we say that it is the area of region such that $$x \geq a$$  and $$0 \leq y \leq f(x)$$

$$A(S)=\displaystyle \int\limits^\infty_a f(x)dx$$

• Note that the area bounded by  $$y=f(x)$$$$x$$  axis,

$$x=a$$  and  $$x=b$$  is given by

$$\displaystyle \int\limits^b_a f(x)dx$$

In this case  $$b \to \infty$$

#### Evaluate the area of the region given by  $$S=\left\{(x,\,y)|\,x\geq 0\, \text{ and}\;0\leq y\leq\;xe^{-2x}\right\}$$

A $$A(S)=\dfrac{1}{4}$$

B $$A(S)=\dfrac{1}{2}$$

C $$A(S)=\dfrac{1}{10}$$

D $$A(S)=5$$

×

$$A(S)\displaystyle=\int\limits^{\infty}_{0}xe^{-2x}dx$$

$$A(S)\displaystyle=\lim\limits_{t\to \infty}\,\int\limits^{t}_{0}xe^{-2x}dx$$

Now consider,

$$\displaystyle \int\limits xe^{-2x}dx$$

$$=\dfrac{xe^{-2x}}{-2}-\displaystyle \int\limits \dfrac{e^{-2x}}{-2}dx$$              (Integration by parts)

$$=x\dfrac{e^{-2x}}{2}+\left(\dfrac{-e^{-2x}}{4}\right)$$

$$=x\dfrac{e^{-2x}}{2}-\dfrac{e^{-2x}}{4}$$

$$\therefore\;\displaystyle \int\limits^{t}_{0} xe^{-2x}dx=\Bigg[\dfrac{xe^{-2x}}{-2}-\dfrac{e^{-2x}}{4}\Bigg]^t_0$$

$$=\left(\dfrac{-te^{-2t}}{2}-\dfrac{e^{-2t}}{4}\right)-\left(\dfrac{-1}{4}\right)$$

$$\therefore\;\displaystyle \int\limits^{\infty}_{0} xe^{-2x}dx=\lim\limits_{t\to \infty} \left(\underbrace{\dfrac{-te^{-2t}}{2}}_{0 \,as\,t\to\infty}+\underbrace{\dfrac{e^{-2t}}{4}}_{0 \,as\,t\to\infty}\right)+\dfrac{1}{4}=\dfrac{1}{4}$$

$$\therefore\;A(S)=\dfrac{1}{4}$$

### Evaluate the area of the region given by  $$S=\left\{(x,\,y)|\,x\geq 0\, \text{ and}\;0\leq y\leq\;xe^{-2x}\right\}$$

A

$$A(S)=\dfrac{1}{4}$$

.

B

$$A(S)=\dfrac{1}{2}$$

C

$$A(S)=\dfrac{1}{10}$$

D

$$A(S)=5$$

Option A is Correct