If $p(t)=\frac{t(t-1) \cdots(t-2019)}{2019!}$, then the value of$$ \int_0^1\left(\frac{1}{t+1}+\frac...

Given,$$ \begin{gathered} \int_0^1\left(\frac{1}{(t+1)}+\frac{1}{(t+2)}+\cdots+\frac{1}{(t+2020)} p(-t-1) d t\right. \\ p(t)=\frac{t(t-1) \ldots(t-2019)}{2019!} \end{gathered} $$The expression for $p(-t-1)$ :$$ \begin{aligned} \Rightarrow p(-t-1) & =\frac{(-t-1)(-t-2) \ldots(-t-2020)}{2019!} \\ p(-t-1) & =\frac{(t+1)(t+2) \ldots(t+2020)}{2019!} \end{aligned} $$Taking log on both sides,$$ \ln (p(-t-1))=\ln (t+1)+\ln (t+2)+\ldots+\ln (t+2020)-\ln (2019!) $$Let $p(-t-1)=y$ and then differentiating,$$ \begin{gathered} \frac{1}{y} \frac{d y}{d t}=\frac{1}{t+1}+\frac{1}{t+2}+\ldots \frac{1}{t+2020} \\ \frac{d y}{d t}=\left[\frac{1}{t+1}+\frac{1}{t+2}+\ldots \frac{1}{t+2020}\right] p(-t-1) \end{gathered} $$On integrating both sides and multiplying with $d t$, we get$$ \int_0^1\left(\frac{1}{t+1}+\frac{1}{t+2}+\cdots+\frac{1}{t+2020}\right) p(-t-1) d t=\int_{y(0)}^{y(1)} d y=[y]_{y(0)}^{y(1)} $$Using$$ y(t)=p(-t-1)=\frac{(t+1)(t+2) \ldots(t+2020)}{2019!} $$We get,$$ y(1)=\frac{2 \cdot 3 \cdot 4 \cdots 2021}{2019!}=\frac{2021!}{2019!}=2021 \times 2020 $$You can similarly find $y(0)=2020$$$ \begin{gathered} =y(1)-y(0)=2021 \times 2020-2020=2020(2021-1) \\ =2020^2 \end{gathered} $$

$F(x)=\int_0^{e^x}\left(t^3+2 t^2-t-2\right) d t$, then for how many real numbers $x$ does $F^{\prim...

Given $ F(x) = \int_0^{e^x} (t^3 + 2t^2 - t - 2) \, dt $.Let's evaluate this integral:$ \begin{aligned} F(x) &= \left[ \frac{t^4}{4} + \frac{2t^3}{3} - \frac{t^2}{2} - 2t \right]_0^{e^x} \\ &= \frac{(e^x)^4}{4} + \frac{2(e^x)^3}{3} - \frac{(e^x)^2}{2} - 2(e^x) \\ &= \frac{e^{4x}}{4} + \frac{2e^{3x}}{3} - \frac{e^{2x}}{2} - 2e^x \end{aligned} $Now, we find the derivative $ F'(x) $:$ \begin{aligned} F'(x) &= \frac{d}{dx} \left( \frac{e^{4x}}{4} + \frac{2e^{3x}}{3} - \frac{e^{2x}}{2} - 2e^x \right) \\ &= e^{4x} + 2e^{3x} - e^{2x} - 2e^x \end{aligned} $Factor out $ e^x $ from the expression:$ F'(x) = e^x \left( e^{3x} + 2e^{2x} - e^x - 2 \right) $Further factorization results in:$ F'(x) = e^x \left( (e^x - 1)(e^x + 2) \right) $For $ F'(x) = 0 $ to hold, one of the factors must be zero. Since $ e^x $ can never be zero or negative for real $ x $, we look at:$ e^x - 1 = 0 $Solve for $ x $:$ e^x = 1 \quad \Rightarrow \quad x = 0 $Thus, there is only one real solution: $ x = 0 $. Therefore, there is a single real number $ x $ for which $ F'(x) = 0 $.

For a natural number $n$, let $C_n$ be the curve in the $X Y$-plane given by $y=x^n$, where $0 \leq$...

To understand the area it is asking us to maximize, observe the following diagram to visualise better:Given,$-n$ is a natural numberThe equation of the area will be given by:$$ \begin{aligned} A(x) & =\int_0^1\left(x^n-x^{n-1}\right) d x \\ & =\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1 \\ & =\frac{1}{n+1}-\frac{1}{n+2} \\ & =\frac{(n+2)-(n+1)}{(n+1)(n+2)} \end{aligned} $$$$ A(x)=\frac{1}{(n+1)(n+2)} $$For the largest value of $A(x)$, we need the smallest denominator (the smallest NATURAL number), i.e., $n=1$$$ A_{\max }=\frac{1}{2 \times 3}=\frac{1}{6} $$

Let $f$ be a continuous function on $[0,1]$ and $F$ be its antiderivative. If $F(0)=1$ and $\int_0^1...

$$ \begin{aligned} &\text { Given, } F(0)=1\\ &\begin{gathered} \int f(x)=F(X) \\ \int_0^1 f(x)=F(1)-F(0)=1 \\ F(1)=1+F(0)=2 \end{gathered} \end{aligned} $$

The value of the integral$$ \int_1^{100} \frac{[x]}{x} d x $$where $[x]$ is the greatest integer les...

The integral is split into segments where $[x]$ is constant. For $n \leq x$$ \int_0^{100} \frac{[x]}{x}=\int_0^1 \frac{0}{x}+\int_1^2 \frac{1}{x}+\int_2^3 \frac{3}{x} \cdots+\int_{99}^{100} \frac{99}{x} $$$\int \frac{1}{x} d x=\ln |x|$, Here $\ln |x|=\ln (x)$ as all values being used are positive$$ \begin{gathered} =0+[\ln (x)]_1^2+[2 \ln (x)]_2^3+[3 \ln (x)]_3^4 \cdots+[99 \ln (x)]_{99}^{100} \\ =[\ln (2)-0]+[2 \ln (3)-2 \ln (2)]+[3 \ln (4)-3 \ln (3)] \cdots+[99 \ln (100)-99 \ln (99)] \end{gathered} $$Rearrange the terms to group similar logarithms.$$ \begin{gathered} =[\ln (2)-2 \ln (2)]+[2 \ln (3)-3 \ln (3)]+[3 \ln (4)-4 \ln (4)] \cdots[98 \ln (99)-99 \ln (99)] \\ +99 \ln (100) \\ =-\ln (2)-\ln (3)-\ln (4)-\ln (5) \cdots-\ln (99)+99 \ln (100) \\ =99 \ln (100)-\ln (99)-\ln (98)-\ln (97) \cdots-\ln (2) \\ =\ln \left(100^{99}\right)-\ln (99!) \\ =\ln \left(\frac{100^{99}}{99!}\right) \end{gathered} $$

Let $f: \mathbf{R} \rightarrow(0, \infty)$ be a continuous decreasing function. Suppose $f(0), \dot{...

Since it is given that $ f(0), f(1), \ldots, f(10) $ form a geometric progression (GP) with a common ratio of $\frac{1}{5}$, we can express the function as:$ f(x) = f(0) \left(\frac{1}{5}\right)^x $To find the integral $\int_0^{10} f(x) \, dx$, we integrate the function:$ \int a^x \, dx = \frac{a^x}{\ln(a)} + C $Substituting our function into the integral:$ \int_0^{10} f(x) \, dx = f(0) \int_0^{10} \left(\frac{1}{5}\right)^x \, dx = f(0) \left[ \frac{\left(\frac{1}{5}\right)^x}{\ln\left(\frac{1}{5}\right)} \right]_0^{10} $Calculate the definite integral:$ = f(0) \cdot \left[ \frac{1}{5}^{10} - 1 \right] \times \frac{1}{-\ln(5)} $Simplifying:$ = \frac{f(0) \left[1 - \frac{1}{5}^{10}\right]}{\ln(5)} $Using a calculator to approximate this result:$ = f(0) \times 0.62 $Therefore, we conclude:$ \int_0^{10} f(x) \, dx \in (0, 2f(0)) $

Let $f:(-1,2) \rightarrow \mathbf{R}$ be a differentiable function such that $f^{\prime}(x)=\frac{2}...

$$ f^{\prime}(x)=\frac{2}{x^2-5} $$On integrating both side wrt $x$,$$ f(x)=\int \frac{2}{x^2-5} d x $$Using the identity,$$ \int \frac{1}{x^2-a^2} d x=\frac{1}{2 a} \ln \left|\frac{x-a}{x+a}\right|+c $$Here, $a=\sqrt{5}$$$ f(x)=2 \int \frac{1}{x^2-5} d x=\frac{2}{2 \sqrt{5}} \ln \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+c $$It is given that, $f(0)=0$$$ \begin{gathered} \rightarrow \frac{1}{\sqrt{5}} \ln |-1|+c=0 \\ c=0 \text { as } \log |-1|=\log 1=0 \\ f(1)=\frac{1}{\sqrt{5}} \ln \left|\frac{1-\sqrt{5}}{1+\sqrt{5}}\right|=\frac{1}{\sqrt{5}} \ln \left|\frac{(1-\sqrt{5})^2}{4}\right| \end{gathered} $$Using a calculator, we get $f(0) \approx-.43$ which lies in $(-\infty, 0)$

Let $I=\int_{e^{-\pi / 2}}^{e^{\pi / 2}}\left(\sin ^2(\log (x))+\sin \left(\log \left(x^2\right)\rig...

$$ \int_{e^{-\pi / 2}}^{e^{\pi / 2}}\left(\sin ^2(\log (x))+\sin (2 \log (x))\right) d x $$Let's use the substitution $t=\log (x)$. Also note, $x=e^t$Then, $d t=\frac{1}{x} d x$ or $d x=x d t=e^t d t$$$ \int_{-\pi / 2}^{\pi / 2}\left(\sin ^2(t)+\sin (2 t)\right) e^t d t $$Note, $\frac{d}{d x}\left(\sin ^2(x)\right)=2 \sin (x) \cos (x)$Now, using $\sin (2 t)=2 \sin (t) \cos (t)$ :$$ \int_{-\pi / 2}^{\pi / 2}\left(\sin ^2(t)+2 \sin (t) \cos (t)\right) e^t d t $$This is present in the form:$\int e^x\left[f(x)+f^{\prime}(x)\right] d x$ which is equal to $\int e^x f(x) d x$ (NCERT XII page 327)So,$$ \int_{-\pi / 2}^{\pi / 2}\left(\sin ^2(t)+2 \sin (t) \cos (t)\right) e^t d t .=\left[\sin ^2(t) e^t\right]_{-\pi / 2}^{\pi / 2}=e^{\pi / 2}-e^{-\pi / 2} $$

Definite Integration - IAT-IISER Maths PYQs

01

PYQ 2020

easy

maths ID: iat-iise

If, then the value of$ $is:

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2019

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2020

02

PYQ 2020

easy

maths ID: iat-iise

, then for how many real numbers does ?

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03

PYQ 2022

easy

maths ID: iat-iise

For a natural number, let be the curve in the -plane given by, where . Let denote the area of the region bounded between and . Then the largest value of is

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04

PYQ 2022

easy

maths ID: iat-iise

Let be a continuous function on and be its antiderivative. If and, then is

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05

PYQ 2022

easy

maths ID: iat-iise

The value of the integral$ [x] x x$, is

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06

PYQ 2023

medium

maths ID: iat-iise

Let be a continuous decreasing function. Suppose are in a geometric progression with common ratio . In which of the following intervals does the value of lie?

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07

PYQ 2023

medium

maths ID: iat-iise

Let be a differentiable function such that and . Then in which of the following intervals does lie?

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08

PYQ 2024

easy

maths ID: iat-iise

Let . What is the value of ?

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PYQ 2025

easy

maths ID: iat-iise

What is the value of ?

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Definite Integration

High-Yield Trend

Questions
9 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Definite Integration

High-Yield Trend

Questions 9 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Questions
9 MCQs