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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.

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.

Illustration Questions

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

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.

Illustration Questions

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.

Illustration Questions

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

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.

Illustration Questions

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.

Illustration Questions

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

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.

Illustration Questions

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

\(\therefore\;I_0=1\)

\(\therefore\;I_2=2I_0=2\)

Illustration Questions

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

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.

Illustration Questions

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

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