Definite Integral as Limit of a sum | ISI QMS | QMA 2019

Try this problem from ISI QMS 2019 exam. It requires knowledge Real Analysis and integral calculus and is based on Definite Integral as Limit of a sum.

Definite integral as Limit of a Sum - ISI QMS (QMB Problem 1a)

Find the value of : $ \displaystyle \lim_{n \to \infty}\big[\frac1n+\frac{1}{n+1}+\ldots + \frac{1}{3n}\big]$

Key Concepts

Real Analysis

Definite Integral

Riemann Sum

Check the Answer

Answer: $\ln 3$

ISI QMS 2019 (QMB Problem 1a)

Secrets in Inequalities

Try with Hints

Putting it into standard form :

$\displaystyle\lim_{n \to \infty} \big[\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{3n}\big]$

$= \displaystyle \lim_{n\to \infty}\big[\frac{1}{n+0}+\frac{1}{n+1}+\ldots+\frac{1}{n+2n}\big]$

$= \displaystyle \lim_{n \to \infty} \displaystyle\sum_{r=0}^{2n} \frac{1}{n+r}$

$= \displaystyle \lim_{n \to \infty} \frac1n\displaystyle\sum_{r=0}^{2n} \frac{1}{1+\frac rn}$

$= \displaystyle \lim_{n \to \infty} \frac1n\displaystyle\sum_{r=0}^{2n} f(\frac rn)$ where $f(x)=\frac{1}{1+x}$

An useful result : Let $f:[a,b]\to \mathbb R$ be integrable on $[a,b]$ and $\{P_n\}$ be a sequence of partitions of $[a,b]$ such that the sequence $\{\parallel P_n\parallel\}$ converges to $0$. Then if $\epsilon >0$ be given, there exists a natural number $k$ such that $|S(P_n, f)-\int_a^b f|<\epsilon \quad \forall n\geq k$ where $S(P,f)$ is a Riemann sum of $f$ corresponding to $P$ and any choice of intermediate points.

i.e., If $f$ be integrable on $[a,b]$ and $\{P_n\}$ be a sequence of partitions of $[a,b]$ such that $\lim\limits_{n \to \infty} \parallel P_n \parallel=0$, then $\lim\limits_{n\to \infty} S(P_n,f)=\int_a^b f$

Let $P_n=(0,\frac1n,\frac2n,\ldots,\frac{2n}{n})$ be a sequence of partition on $[0,2]$ dividing it into $2n$ sub-intervals of equal length $\frac1n$.

Also $\lim \parallel P_n\parallel=\lim \frac1n=0$.

Let us choose $\xi_r=\frac rn,\quad r=1,2,3,\ldots,2n$

Then the Riemann sum for $f$ on the interval $[0,2]$ corresponding to the partition $P_n$ and chosen intermediate points $\xi_r$

$S(P_n,f)=\frac1n\displaystyle\sum_{r=1}^{2n} f(\frac rn)$

As $f$ is continuous on $[0,2]$, $f$ is integrable on $[0,2]$.

Now can you use the above result to reach to the solution ?

Using the result

$\lim\limits_{n\to\infty} \frac1n\displaystyle\sum_{r=1}^{2n} f(\frac rn) = \displaystyle\int_0^2 f(x) \mathrm d x$

$= \displaystyle\int_0^2 \frac{ \mathrm d x }{1+x} $

$=\ln(1+x)\bigg|_0^2= \ln 3$ [Ans]

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Definite Integral | IIT JAM 2018 | Problem 4

Try this beautiful problem from IIT JAM 2018 which requires knowledge of the properties of Definite integral.

Properties of Definite Integral -IIT JAM2018 (Problem 4)

Let $a$ be positive real number. If $f$ is a continuous and even function defined on the interval $[-a,a]$, then $\displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x$ is equal to :-

  • $\displaystyle\int_0^a f(x) \mathrm d x$
  • $2\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x$
  • $2\displaystyle\int_0^a f(x) \mathrm d x$
  • $2a\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x$

Key Concepts

Definite Integral

Properties of definite Integral

Even function / Odd function

Check the Answer

Answer: $ \displaystyle\int_0^a f(x) \mathrm d x $

IIT JAM 2018, Problem 4

Definite and Integral calculus : R Courant

Try with Hints

In this first I will give you the properties we need to solve this problem :

Property 1 : $\displaystyle\int_a^b f(x) \mathrm d x = \displaystyle\int_a^b f(a+b-x) \mathrm d x $

[Where $f$ is continuous on $[a,b]$]

Property 2 : If $f$ is an even function i.e., $f(x)=f(-x)$ then

$ \displaystyle\int_{-a}^{a} f(x) \mathrm d x = 2 \displaystyle\int_{0}^{a} f(x) \mathrm d x $

Can you drive it from here !!!! Give it a try !!!

Let $I=\displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x \quad \ldots (i)$

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(a-a-x)}{1+e^{(a-a-x)}} \mathrm d x $

[Since, $f$ is continuous then $\displaystyle\int_{a}^b f(x) \mathrm{d}x = \displaystyle\int_{a}^b f(a+b-x) \mathrm{d} x $]

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(-x)}{1+e^{-x}} \mathrm d x$

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(x)}{1+\frac{1}{e^x}} \mathrm d x$ [Since $f(x)$ is even]

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{e^x.f(x)}{1+e^{x}} \mathrm d x \quad \ldots (ii) $

Adding $(i)$ and $(ii)$ we can get some interesting result !!!

Adding $(i)$ and $(ii)$ we get ,

$2I= \displaystyle\int_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm d x + \displaystyle\int_{-a}^a \frac{e^x .f(x)}{1+e^{x}} \mathrm d x$

$\Rightarrow 2I = \displaystyle\int_{-a}^a \frac{[f(x)+e^x.f(x)]}{1+e^{x}} \mathrm d x$

$\Rightarrow 2I = \displaystyle\int_{-a}^a \frac{f(x)[1+e^x]}{[1+e^{x}]}$

$ \Rightarrow 2I= \displaystyle\int_{-a}^a f(x) \mathrm d x$

$\Rightarrow 2I = 2\displaystyle\int_0^a f(x) \mathrm d x $ [Since $f(x) $ is even ]

$\Rightarrow I = \displaystyle\int_0^a f(x) \mathrm d x $ [ANS]

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Triple Integral | IIT JAM 2016 | Question 15

Question 15 - Triple Integral (IIT JAM 2016)

If the triple integral over the region bounded by the planes $2x+y+z=4$ $x=0$ $y=0$ $z=0$ is given by $\int\limits_0^2\int\limits_0^{\lambda(x)}\int\limits_0^{\mu(x,y)}\mathrm d z\mathrm d y\mathrm d x$ then the function $\lambda(x)-\mu(x,y)$ is

  • $x+y$
  • $x-y$
  • $x$
  • $y$

Key Concepts

Real Analysis

Integral Calculus

Triple Integral

Check the Answer

Answer: $\textbf{(B)} \quad y$

IIT JAM 2016, Question No. 15

Differential and Integral Calculus: R Courant

Try with Hints

Here we are given with triple integral over the region bounded by the planes $2x+y+z=4, x=0, y=0$ and $z=0$

Now we our aim here is to find $\lambda (x) $ and $\mu(x,y)$. Now we will approach this problem by find the volume of $(x,y,z)$ based on $2x+y+z=4$ can you do this ??? (With the given information x=0, y=0, z=0)

$2x+y+z=4$

$\Rightarrow z=4-2x-y$

$\Rightarrow 2x+y=4$ [as $z=0$]

$\Rightarrow y=4-2x$

Again,

$2x+y+z=4$

Now as $y=z=0$ we have $2x=4$

Therefore $x=2$

Now can you use this to move forward with this problem ?

So our triple integral become,

$\int_0^2\int_0^{(4-2x)}\int_0^{(4-2x-y)}\mathrm d z\mathrm d y\mathrm dx$

On compairing $\lambda(x)=4-2x$ and $\mu(x,y)=4-2x-y$

Therefore $\lambda(x)-\mu(x,y)=4-2x-4+2x+y=y$ (ANS)

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Relation Mapping (IIT JAM 2014)

Question 33 - Relation-Mapping (IIT JAM 2014)

A beautiful problem involving the concept of relation-mapping from IIT JAM 2014.

Let $f:(0,\infty)\to \mathbb R$ be a differentiable function such that $f'(x^2)=1-x^3$ for all $x>0$ and $f(1)=0$. Then $f(4)$ equals

  • $\frac{-47}{5}$
  • $\frac{-47}{10}$
  • $\frac{-16}{5}$
  • $\frac{-8}{5}$

Key Concepts

Relation/Mapping

Differentiation

Integration

Check the Answer

Answer: (A) $ \frac{-47}{5}$

Try with Hints

The above problem can be done in many ways we will try to solve this by the simplest method.

Now, as the function is given as $f'(x^2)=1-x^3$

So first try to change this $x^3$ into $x^2$. Try this. It's very easy !!!

To change $x^3$ into $x^2$ we can easily do

$f'(x^2)=1-(x^2)^{\frac32}$

Now we have to find the value of $f(4)$ so we have to change the second degree term, i.e., $x^2$ into some linear form. Can you cook this up ???

Let us assume $x^2=y$

i.e., $f'(y)=1-y^{\frac32}$

Now you know from previous knowledge that integration is also known as anti-derivative. So $f'(y)$ can be changed into $f(y)$ by integrating it with respect to $y$. Try to do this integration and we are half way done !!!

On integrating both side w.r.t $y$ we get :

$f(y)=y-\frac25y^{\frac52}+c$, (where $c$ is a integrating constant.)

Now we find the value to $c$

We know $f(1)=0$

$\Rightarrow c=-\frac35$

i.e., $f(y)=y-\frac25y^{\frac52}-\frac35$

Can you find the answer now ?

Now simply, putting $y=4$

we get $f(4)=4-\frac25(4)^{\frac52}-\frac35 \\=4-\frac{64}{5}-\frac35 \\= \frac{20-67}{5} \\= -\frac{47}{5}$

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Line Integral : IIT JAM 2018 Question Number 20

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What are we learning ?

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Competency in Focus: Application of Calculus This is problem from IIT JAM 2018 is based on calculation of Line integration of a vector point function along a closed curve.[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/02/IIT_2018_JAM_20.png" align="center" force_fullwidth="on" _builder_version="4.2.2" min_height="388px" height="198px" max_height="207px"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]If  $\vec{F}(x,y)=(3x-8y)\widehat{i}+(4y-6xy)\widehat{j}$ for $(x,y) \in \mathbb{R}^2$, then $\oint \vec{F}. \mathrm d \vec{r} $, where $C$ is the boundary of triangular region bounded by the lines $x=0,y=0,\quad \textbf{and} \quad x+y=1$ oriented in the anti clock wise direction is    $(A)\quad \frac{5}{2} \qquad (B)\quad 3 \qquad (C)\quad 4 \qquad (D)\quad 5 \qquad $[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.2.2" width="100%" max_width="1024px" max_width_tablet="" max_width_phone="" max_width_last_edited="on|phone"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.2.2" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.2.2"]IIT JAM 2018, Question number 20[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.2.2" open="off"]Calculation of line integral of a vector point function along a closed curve.[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.2.2" open="off"]6/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.2.2" open="off"]VECTOR ANALYSIS: Schaum’s Outlines series[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|0px|20px||" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints 

[/et_pb_text][et_pb_tabs _builder_version="4.2.2" hover_enabled="0"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]Suppose we have vectors $\vec{u}=x_{1}\widehat{i}+y_{1}\widehat{j}+z_{1}\widehat{k}$ $\vec{v}=x_{2}\widehat{i}+y_{2}\widehat{j}+z_{2}\widehat{k}$ Then $\vec{u}.\vec{v}=x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}$ Now in the given context $\vec{F} (x,y)=(3x-8y)\widehat{i}+(4y-6xy)\widehat{j}$ $\mathrm d \vec{r}=\mathrm d x\widehat{i}+\mathrm d y\widehat{j}$ Then $\oint_{c} \vec{F}. \mathrm d \vec{r}=\oint_c (3x-8y) \mathrm d x+(4y-6xy)\mathrm d y$ Now we have to chose the proper path to complete the line integral. Do you want to try from here?  [/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]The region is bounded by $x=0=y$ & $x+y=1$ i.e., Let us divide the segment into 3 parts where $I_1 : y=0 ; \quad I_2 : x+y=1; \quad I_3 : x=0 $ i.e., $\oint_c \vec{F}. \mathrm d \vec{r} = \oint_{I_1} \vec{F}. \mathrm d \vec{r} + \oint_{I_2} \vec{F}. \mathrm d \vec{r}+\oint_{I_3} \vec{F}. \mathrm d \vec{r}$   Let us calculate $\oint_{I_1} \vec{F}. \mathrm d \vec{r}$ & $\oint_{I_3} \vec{F}. \mathrm d \vec{r}$ and I'll keep $\oint_{I_2} \vec{F}. \mathrm d \vec{r}$  for your try by the end of this hint. $\oint_{I_1} \vec{F}. \mathrm d \vec{r}=\int_0^1 3x \mathrm d x$ [We have put $y=0$ so $\mathrm d y=0$ , Hence the limit will be on $x$] $=\frac{3}{2}$ $\oint_{I_3} \vec{F}. \mathrm d \vec{r}=\int_1^0 4y \mathrm d y$ [We have put $x=0$ so, $\mathrm d x=0$ therefore the limit will be on $y$; Observe the anticlockwise direction to understand the limit] $=[\frac{4y}{2}]_1^0$ $=-2$[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2" hover_enabled="0"]On $I_2$  observe that we can transform $y$ in terms of $x$ i.e., $y=1-x$ [We can also use parameterization, which is basically the same method but I prefer this one to make the calculation faster] $y=1-x $ i.e., $\mathrm d y=-\mathrm d x$ now,  $\int_{I_2} \vec{F}.\mathrm d \vec{r}= \int_{I_2}(3x-8y)\mathrm d x+\int_{I_2}(4y-6xy)\mathrm d y$ $=\int_1^0(3x-8+8x)\mathrm d x+\int_1^0(4-4x-6x(1-x))(-\mathrm d x)$ [observe the anticlock wise direction limit] $=\int_1^0(11x-8)\mathrm d x +\int_1^0(10x-6x^2-4)\mathrm dx$ $=[\frac{11x^2}{2}-8x+\frac{10x^2}{2}-\frac{6x^3}{3}-4x]_0^1$ $=-\frac{11}{2}+8-5+2+4=\frac{7}{2}$ Hence our answer would be $\frac{3}{2}-2+\frac{7}{2}=3$ [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Similar Problems

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Volume of revolution : IIT JAM 2018 Question Number 19

Competency in Focus: Application of Calculus (Volume of Revolution)
This problem is from IIT JAM 2018 (Question number 19) and is based on the calculation of the volume of revolution.

Consider the region (D) on (yz) plane and bounded by the line (y=\frac{1}{2}) and the curve (y^{2}+z^{2}=1) where (y\geq0). If the region (D) is revolved about the (z-)axis then the volume of the resulting solid is
$ (A)\frac{\pi}{\sqrt{3}}\qquad$

$(B)\frac{2\pi}{\sqrt{3}} \qquad$

$(C)\frac{\pi\sqrt{3}}{2}\qquad$

$(D)\pi\sqrt{3} $

Do you really need a hint? Try it first!

Hint 1

Imagine that we have a portion of a curve.
\(y=f(x)\) from \(x=a\) to \(x=b\).
In the \(xy-plane\) we revolve it around a straight line - \(x\)-axis.
The result is called solid of revolution
Here in our next hint we will find techniques to calculate the volumes of solid of revolution.

Hint 2

Calculate the volumes of solid of revolution

Let's construct a narrow rectangle of base width \(\mathrm d x\) and height \(f(x)\) sitting under the curve. When this rectangle is revolved around the \(x-\text{axis}\). We get a disk whose radius is \(f(x)\) and height is \(\mathrm d x\).
The volume of this disk \(\mathrm d V=\pi[f(x)]^{2}\mathrm d x\).
So the total volume of the solid is 
\(V=\int_a^b \mathrm d V=\int_a^b \pi[f(x)]^{2}\mathrm d x\).
If the curve revolved around the vertical line (such as \(y\)-axis), then horizontal disks are used. If the curve can be solved for (x) in terms of (y). \(x=g(y)\) then the formula would be.
\(V=\int_a^b[g(y)]^2\mathrm d y\).
Now can you guess if the region between two curves is revolved around the axis.

HINT 3

Sometimes the region between two curves is revolved around the axis and a gap is created between the solid and the axis. A rectangle within the rotated region will become a disk with a hole in it, also known as washer. If the rectangle is vertical and extends from the curve \(y=g(x)\) up to the curve \(y=f(x)\), then when it is rotated around (x)- axis, it will result in a washer with volume equal to
\(\mathrm d V=\pi{[f(x)]^{2}-[g(x)]^{2}}\mathrm d x\).
Which gives us 
\(V=\int_a^b \pi {[f(x)]^{2}-[g(x)]^{2}}\mathrm d x\).

HINT 4

Now in the given problem replace (x) by (z) , in the above discussion
\(y=f(z)=\sqrt{1-z^{2}}, y=g(z)=\frac{1}{2}\)
and the line \(y=\frac{1}{2}\) intersects the curve at ( \(\frac{-\sqrt{3}}{2},\frac{1}{2}\)) , ( \(\frac{\sqrt{3}}{2},\frac{1}{2}\)) in terms of (\(z,x)\) co-ordinate.
and hence the volume is 
\(V=\pi \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}[(1-z^{2})-\frac{1}{4})]\mathrm d z\).
\(=\pi[\frac{3z}{4}-\frac{z^{3}}{3}]_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\)
\(=\pi(\frac{3\sqrt{3}}{8}-\frac{3\sqrt{3}}{24})\times 2\)
\(=\pi \frac{3\sqrt{3} \times 2 \times 2}{24}\)
\(=\frac{\pi \sqrt{3}}{2}\)

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Integral Calculus by Gorakh Prasad

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Integration of nonnegative function: TIFR 2018 Part A, Problem 4

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.26.6" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let $latex f$ be a nonnegative continuous function on $latex \Bbb R$ s.t $latex \int_0^{\infty}f(t)dt$ is finite then $latex \lim_{x \to \infty}f(x)=0$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.26.6" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.26.6"]TIFR 2018 Part A, Problem 4 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.26.6" open="off"]Analysis [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.26.6" open="off"]Hard [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.26.6" open="off"]Real Analysis; Bartle and Sherbert [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.26.6"]Can you think of a function such that it has finite area but as $latex x$ goes to infinity, $latex f(x)$ goes to infinity?

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.26.6"]

Try to sketch the above function. Picture[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.26.6"]

See if the $latex lim f(x)$ exists then it must be zero right? but what if the limit fails to exist and the integral still gives you finite value? Hence, this is the case drawn in the above picture. Hence the answer is false generally.

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Watch the video

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Connected Program at Cheenta

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Similar Problems

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TIFR 2014 Problem 21 Solution - Determining function


TIFR 2014 Problem 21 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:

Let $f:[0,1]\to [0,\infty )$ be continuous. Suppose $\int_{0}^{x} f(t)dt \ge f(x)$ for all $x\in [0,1]$.

Then

A. no such function exists

B. there are infinitely many such functions

C. there is only one such function

D. there are exactly two such functions

Discussion:

Basically, the question is to find out how many such functions can exist.

Let $F(x)=\int_{0}^{x} f(t)dt $. That is, $F(x)$ is the indefinite integral of $f$.

We know from the fundamental theorem of calculus that:

$F'(x)=f(x)$ .

So we have $F'(x) \ge F(x) $ for all $x\in [0,1] $.

We will manage this equation as we do in case of differential equations. Except that, we have an inequality here.

$F'(x)-F(x) \ge 0$ for all $x\in [0,1]$.

Multiplying both sides by $e^{-x}$ the inequality remain unchanged. This is because $e^{-x} >0$.

$e^{-x}F'(x)-e^{-x}F(x) \ge 0$ for all $x\in [0,1] $.

Now, $e^{-x}F(x)' = e^{-x}F(x) -e^{-x}F'(x)$.

So we have $e^{-x}F(x))' \le 0$ for all $x\in [0,1]$.

Therefore, by taking integral from o to y and by using fundamental theorem of calculus:

$e^{-y}F(y)-e^{0}F(0) \le 0$ for all $y\in [0,1]$.

i.e, $e^{-x}F(x) \le F(0)$ for all $x\in [0,1]$.

Also, note that by definition of (F), (F(0)=0). So we have

$e^{-x}F(x) \le 0$ for all $x\in [0,1]$.

But, since $e^{-x} >0$ for all $x\in [0,1]$, we have $F'(x) \le 0$  for all $x\in [0,1]$.

So far, we have not used the fact that $f$ is a non-negative function. Now we use it. Since $f(x) \ge 0$ for all $x\in [0,1]$, therefore by monotonicity of the integral, $F$ is an increasing function. This means $F'(x) \ge 0$ for all $x\in [0,1]$.

By the two inequalities obtained above, we get $F'(x) = 0$  for all $x\in [0,1]$.

By the fundamental theorem (again!) we get $f(x)=F'(x)=0$  for all $x\in [0,1]$.

So there is only one such $f$ namely the constant function $f=0$.

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TIFR 2014 Problem 7 Solution -Limit of Integration


TIFR 2014 Problem 7 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Real and Complex Analysis by Walter Rudin. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem

Let (f_n(x); n\ge 1) be a sequence of continuous nonnegative functions on ([0,1]) such that

( \lim_{n\to\infty} \int_{0}^{1}f_n(x)dx = 0 )

Which of the following statements is always correct?

A. (f_n \to 0) uniformly of ([0,1])

B. (f_n) may not converge uniformly but converges to (0) point-wise.

C. (f_n) will converge point-wise and the limit may be non-zero.

D. (f_n) is not guaranteed to have a point-wise limit.

Discussion:

We start by a very well known example: (g_n(x)=x^n).

(  \int_{0}^{1}g_n(x)dx= \frac{1}{n+1}\to 0 \int_{0}^{1}f_n(x)dx ) as (n\to \infty).

We know (g_n) does not converge uniformly on ([0,1]) because the limit is (1) at (x=1) and (0) everywhere else so we have a non-continuous limit.

So straight-away A,B are false. Question is now whether at all the sequence has to have a point-wise limit or not.

For this, we take our hint from (g_n) and construct (f_n(x)= \sqrt{n}x^n ).

Then ( \int_{0}^{1}f_n(x)dx= \sqrt{n}\frac{1}{n+1}\to 0 ) as ( n\to \infty).

But look at (f_n(1)= \sqrt{n} ). Therefore, (f_n) does not converge at the point (x=1).

So option (D) i.e., (f_n) is not guaranteed to have a point-wise limit. is true.

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