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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.

Illustration Questions

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.

Illustration Questions

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.

Illustration Questions

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.

Illustration Questions

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.

Illustration Questions

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.

Illustration Questions

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  \(b \to \infty\).

Illustration Questions

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.

image

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

image

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

\(A(S)=2\)

.

image
B

\(A(S)=10\)

image
C

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

image
D

\(A(S)=1\)

image

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  \(b \to \infty\)

Illustration Questions

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

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