Get Answers to all your Questions

Filter By

Clear Filter

All Questions

Given : $f(x)=\left\{\begin{matrix} x, &0\leq x< \frac{1}{2} \\ \frac{1}{2}, &x=\frac{1}{2} \\ 1-x, &\frac{1}{2} < x\leq 1 \end{matrix}\right.$ and $g(x)=\left ( x-\frac{1}{2} \right )^{2},\: x\epsilon \textbf{R}.$ Then the area (in sq. units) of the region bounded b the curves, $y=f(x)$ and $y=g(x)$ between the lines, $2x=1\: \: \: and\: \: \: 2x=\sqrt{3},$ is :
Option: 1 $\frac{\sqrt{3}}{4}-\frac{1}{3}$

Option: 2 $\frac{1}{3}+\frac{\sqrt{3}}{4}$

Option: 3 $\frac{1}{2}+\frac{\sqrt{3}}{4}$

Option: 4 $\frac{1}{2}-\frac{\sqrt{3}}{4}$

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves y = f(x), y = g(x) and intersect each other in the interval [a, b]

First find the point of intersection of these curves y = f(x) and y = g(x) , let the point of intersection be x = c

Area of the shaded region

$=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x$

When two curves intersects more than one point

rea bounded by the curves y=f(x), y=g(x) and intersect each other at three points at x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

$\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx$

Required area = Area of trapezium ABCD - $\int_{1/2}^{\sqrt3/2}\left ( x-\frac{1}{2} \right )^2dx$

$\\=\frac{1}{2}\left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\left(x-\frac{1}{2}\right)^{3}\right)_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\\=\frac{\sqrt3}{4}-\frac{1}{3}$

Correct Option (1)

View Full Answer(1)

Posted by

avinash.dongre

Let a function $f:\left [ 0,5 \right ]\rightarrow \textbf{R}$ be continuous, $f\left ( 1 \right )=3$ and F be defined as : $F(x)=\int_{1}^{x}t^{2}g\left ( t \right )dt$ , where $g(t)=\int_{1}^{t}f\left ( u \right )du.$ Then for the function F, the point $x=1$ is :
Option: 1 a point of infection.
Option: 2 a point of local maxima.
Option: 3 a point of local minima.
Option: 4 not a critical point.

Integration as Reverse Process of Differentiation -

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given.

For example,

$\\\mathrm{\frac{d}{dx}(\sin x)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2 \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x \right )=e^x}$

In the above example, the function cos x is the derived function of sin x. We say that sin x is an anti derivative (or an integral) of cos x. Similarly, x² and e^x are the anti derivatives (or integrals) of 2x and e^x respectively.

Also note that the derivative of a constant (C) is zero. So we can write the above examples as:

$\\\mathrm{\frac{d}{dx}(\sin x+c)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2+c \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x +c\right )=e^x}$

Thus, anti derivatives (or integrals) of the above functions are not unique. Actually, there exist infinitely many anti derivatives of each of these functions which can be obtained by selecting C arbitrarily from the set of real numbers.

For this reason C is referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function.

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx.

By definition,

$\\\mathrm{\int f(x)dx=F(x)+c,\;\;\;where\;\;F'(x)=f(x)\;\;and\;\;'c' \;is\;constant.}$

Maxima and Minima of a Function -

Maxima and Minima of a Function

Let y = f(x) be a real function defined at x = a. Then the function f(x) is said to have a maximum value at x = a, if f(x) ≤ f(a) ∀ a ≥∈ R.

And also the function f(x) is said to have a minimum value at x = a, if f(x) ≥ f(a) ∀ a ∈ R

Concept of Local Maxima and Local Minima

The function f(x) is said to have a maximum (or we say that f(x) attains a maximum) at a point ‘a’ if the value of f(x) at ‘a’ is greater than its values for all x in a small neighborhood of ‘a’ .

In other words, f(x) has a maximum at x = ‘a’, if f(a + h) ≤ f(a) and f(a - h) ≤ f(a), where h ≥ 0 (very small quantity).

The function f(x) is said to have a minimum (or we say that f(x) attains a minimum) at a point ‘b’ if the value of f(x) at ‘b’ is less than its values for all x in a small neighborhood of ‘b’ .

In other words, f(x) has a maximum at x = ‘b’, if f(a + h) ≥ f(a) and f(a - h) ≥ f(a), where h ≥ 0 (very small quantity).

$\\ \begin{aligned} &\mathrm{F}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \mathrm{g}(\mathrm{x})\\ &\Rightarrow \mathrm{F}^{\prime}(1)=1.\;\;g(1)=0\;\;\;\;\;\ldots(1)\;\;\;\;\;\;\;\;(\because\;g(1)=0)\\ &\text { Now }(\mathrm{x})=2 \mathrm{xg}(\mathrm{x})+\mathrm{x}^{2} \mathrm{g}^{\prime}(\mathrm{x}) \end{aligned}$

$\\ \begin{aligned} &\Rightarrow F^{\prime \prime}(x)=2 x g(x)+x^{2} f(x)\;\;\;\;(\because\;g'(x)=f(x))\\ &\Rightarrow F^{\prime \prime}(1)=0+1 \times 3\\ &\Rightarrow F^{\prime \prime}(1)=3\;\;\;\;\;\;\;\;\;\;\ldots (2) \end{aligned}\\\text{From (1) and (2) F(x) has local minimum at x = 1 }$

Correct Option (1)

View Full Answer(1)

Posted by

avinash.dongre

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

The value of $\int_{0}^{2x}\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx$ is equal to :
Option: 1 $2\pi$

Option: 2 $4\pi$

Option: 3 $2\pi ^{2}$

Option: 4 $\pi ^{2}$

Properties of the Definite Integral (Part 2) - King's Property -

Property 4 (King's Property)

This is one of the most important properties of definite integration.

$\\\mathbf{\int_{a}^{b}f(x)\;dx=\int_{a}^{b}f(a+b-x)\;dx}$

Application of Periodic Properties in Definite Integration -

Property 9

If f(x) is a periodic function with period T, then the area under f(x) for n periods would be n times the area under f(x) for one period, i.e.

$\mathbf{\int_{0}^{n T} f(x) d x=n \int_{0}^{T} f(x) d x}$

$\\I=\int _0^{2\pi }\frac{x\sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=\int _0^{2\pi }\frac{(2\pi-x)\sin ^8(2\pi-x)\:}{\sin ^8(2\pi-x)+\cos ^8(2\pi-x)\:\:}dx\\2I=\int _0^{2\pi }\frac{2\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=\int _0^{2\pi }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx$

$\\I=\int _0^{2\pi }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=4\int _0^{\pi/2 }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=4\int _0^{\pi/2 }\frac{\pi \sin ^8(\pi/2-x)\:}{\sin ^8(\pi/2-x)+\cos ^8(\pi/2-x)\:\:}dx\\I=2\int_{0}^{\pi/2}\pi dx\\I=\pi^2$

View Full Answer(1)

Posted by

avinash.dongre

$\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}$ is equal to
Option: 1 $\frac{18}{e^{4}}$
Option: 2 $\frac{27}{e^{2}}$
Option: 3 $3\log ^{3-2}$
Option: 4 $3\log ^{3-2}$

As we learnt

Walli's Method -

Definite integral by first principle

$\int_{a}^{b}f(x)dx= \left ( b-a \right )\lim_{n \to \infty }\frac{1}{n}\left [ f(a) +f(a+h)+f(a+2h)....\right ]$

where

$h=\frac{b-a}{n}$

- wherein

$y=\lim_{n\rightarrow \infty }\left [ \frac{(n+1)(n+2)...(n+2n)}{n^{2n}} \right ]^{1/n}$

$=\lim_{n\rightarrow \infty }\left [ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right ) ...\left ( 1+\frac{2n}{n} \right )\right ]^{1/n}$

$\Rightarrow \log y=\lim_{n\rightarrow \infty }\frac{1}{n}\left [ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right ) ...\left ( 1+\frac{2n}{n} \right )\right ]$

$=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{2n}\log\left ( 1+\frac{r}{n} \right )$

$=\int_{0}^{2}\log\left ( 1+x \right )dx$

$=\left [ \log(1+x).x \right ]-\int_{0}^{2}\frac{1}{1+x}.xdx$

$= \left [ x\log(1+x)-x +\log(1+x)\right ]^{2}_{0}$

$=2\log 3-2+\log 2$

$\Rightarrow y=\frac{27}{e^{2}}$

View Full Answer(1)

Posted by

avinash.dongre

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

If $\int \frac{\cos \: x\: dx}{\sin ^{3}x(1+\sin ^{6}x)^{2/3}}= f(x)(1+ \sin ^{6}x)^{1/\lambda }+c$ where c is a constant of integration, then $\lambda f\left ( \frac{\pi }{3} \right )$ is equal to :
Option: 1 $-\frac{9}{8}$
Option: 2 $\frac{9}{8}$
Option: 3 $2$
Option: 4 $-2$

Integration Using Substitution -

The method of substitution is one of the basic methods for calculating indefinite integrals.

Substitution - change of variable

$\\\mathrm{To\;solve\;the\;integrate\;of\;the\;form}\\\\\mathrm{I=\int f\left ( g(x) \right )\cdot g'(x)\;dx,\;}\\\\\mathrm{\;where\;g(x)\;is\;continuously\;differentiable\;function.}\\\mathrm{put\;\;g(x)=t,\;\;g'(x)\;dx=dt}\\\mathrm{After\;substitution,\;we\;get\;\;\int f(t)\;dt.}\\\text{Evalute this integration and substitute back the value of }t.$

$\\\sin x=t \quad\Rightarrow \cos x d x=d t\\\\\int \frac{d t}{t^{2}\left(1+t^{6}\right)^{2 / 3}}=\int \frac{d t}{t^{3} t^{4}\left(\frac{1}{t^6}+1\right)^{2/3}}\\\\u=\frac{1}{t^{6}}+1 \Rightarrow d u=-\frac{6}{t^{7}} d t\\\\\frac{d u}{-6}=\frac{d t}{t^{7}}\\$

$\\=\int\frac{du}{-6u^{2/3}}=-\frac{1}{2 }u^{1/3}+C\\\\=-\frac{1}{2}\left ( \frac{1}{t^6}+1 \right )^{1/3}+C\\\\=-\frac{1}{2}\left ( \frac{(1+\sin^6x)^{1/3}}{\sin^2x} \right )\\\\f(x)=-\frac{1}{2}\frac{1}{\sin^2x}\;\;\;\lambda=3\\\\\lambda f(\pi/3)=-2$

Correct option (4)

View Full Answer(1)

Posted by

Kuldeep Maurya

Let, $f(x)=\left ( \sin (\tan ^{-1}x)+\sin (\cot ^{-1}x) \right )^{2}-1,$ $\left | x \right |>1.\; If\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left ( \sin ^{-1}(f(x)) \right )$ and $y(\sqrt{3})=\frac{\pi }{6},$ then $y(-\sqrt{3})$ is equal to :
Option: 1 $\frac{\pi }{3}$
Option: 2 $\frac{2\pi }{3}$
Option: 3 $-\frac{\pi }{6}$
Option: 4 $\frac{5\pi }{6}$

Integration as Reverse Process of Differentiation -

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given.

For example,

$\\\mathrm{\frac{d}{dx}(\sin x)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2 \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x \right )=e^x}$

Also note that the derivative of a constant (C) is zero. So we can write the above examples as:

$\\\mathrm{\frac{d}{dx}(\sin x+c)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2+c \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x +c\right )=e^x}$

For this reason C is referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function.

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx.

By definition,

$\\\mathrm{\int f(x)dx=F(x)+c,\;\;\;where\;\;F'(x)=f(x)\;\;and\;\;'c' \;is\;constant.}$

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

$\\\mathrm{\sin^2\mathit{t}+\cos^2\mathit{t}=1}\\\mathrm{1+\tan^2\mathit{t}=\sec^2\mathit{t}}\\\mathrm1+{\cot^2\mathit{t}=\csc^2\mathit{t}}\\\mathrm{\tan \mathit{t}=\frac{\sin \mathit{t}}{\cos \mathit{t}},\;\;\cot \mathit{t}=\frac{\cos\mathit{t}}{\sin\mathit{t}}}$

Principal Value of function f-1 (f (x)) -

Principal Value of function f^-1 (f (x))

$\begin{array} {l}\mathrm{1.\;\;\sin^{-1}(\sin (\theta))=\theta} \quad\quad\quad \;\mathrm{for\;all\;\theta\in[-\pi/2,\pi/2] }\\\mathrm{2.\;\;\cos^{-1}(\cos(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[0,\pi]}\\\mathrm{3.\;\;\tan^{-1}(\tan(\theta))=\theta} \;\;\;\quad\quad \mathrm{for\;all\;\theta\in(-\pi/2,\pi/2)}\\\mathrm{4.\;\;\cot^{-1}(\cot(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in(0,\pi)} \\\mathrm{5.\;\;\sec^{-1}(\sec(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[0,\pi]-\left \{ \pi/2 \right \}}\\\mathrm{6.\;\;\csc^{-1}(\csc(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[-\pi/2,\pi/2]-\left \{ 0 \right \}}\end{array}$

$\\y=\frac{1}{2} \sin ^{-1}(f(x))+c\\\\\begin{array}{l}{\text {Let } \theta=\tan ^{-1} x \Rightarrow \cot ^{-1} x=\frac{\pi}{2}-\theta} \\\\ {f(x)=\left\{\sin \theta+\sin \left(\frac{\pi}{2}-\theta\right)\right\}^{2}-1}\end{array}\\\\\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1=\sin2\theta$

$\\y=\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \tan ^{-1} x\right)\right)+C\\\\\frac{\pi}{6}=\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \frac{\pi}{3}\right)\right)+C\\\\\frac{\pi}{6}=\frac{1}{2} \frac{\pi}{3}+C\Rightarrow C=0\\\\y=\frac{1}{2} \sin ^{-1}\left(\sin \left(-2 \frac{\pi}{3}\right)\right)=-\frac{\pi}{6}$

Correct Option (4)

View Full Answer(1)

Posted by

Kuldeep Maurya

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

For $a>0,$ let the curves $C_{1}:y^{2}=ax$ and $C_{2}:x^{2}=ay$ intersect at origin O and a point P. Let the line $x=b(0<b<a)$ intersect the chord OP and the x-axis at points Q and R,respectively. If the line $x=b$ bisects the are bounded by the curves, $C_{1}$ and $C_{2'}$ and the area of $\Delta OQR=\frac{1}{2}$ , then 'a' satisfies the equation :

Option: 1 $x^{6}-12x^{3}+4=0$
Option: 2 $x^{6}-12x^{3}-4=0$
Option: 3 $x^{6}+6x^{3}-4=0$
Option: 4 $x^{6}-6x^{3}+4=0$

Parabola -

Parabola

A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix in the plane.

Standard equation of a parabola

Let focus of parabola is S(a, 0) and directrix be x + a = 0, and axis as x-axis

P(x, y) is any point on the parabola.

Now, from the definition of the parabola,
$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;SP=PM}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;SP^2=PM^2}\\\mathrm{\Rightarrow \;\;\;\;\;(x-a)^{2}+(y-0)^{2}=(x+a)^{2}}\\\mathrm{\Rightarrow \;\;\;\;\;y^2=4ax}$

which is the required equation of a standard parabola

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves y = f(x), y = g(x) and intersect each other in the interval [a, b]

First find the point of intersection of these curves y = f(x) and y = g(x) , let the point of intersection be x = c

Area of the shaded region

$=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x$

When two curves intersects more than one point

rea bounded by the curves y=f(x), y=g(x) and intersect each other at three points at x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

$\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx$

$\begin{array}{l}{\frac{1}{2} (b \times b)=\frac{1}{2}} \\ {b=1}\end{array}$

$\int_{0}^{1}\left(\sqrt{a} \sqrt{x}-\frac{x^{2}}{a}\right) d x=\frac{a^2}{6} \text{ by property of parabola}$

By solving above you will get

$a^{6}-12 a^{3}+4=0$

Correct option (1)

View Full Answer(1)

Posted by

Kuldeep Maurya

If $\int \frac{x+1}{\sqrt{2x-1}}dx=f(x)\sqrt{2x-1}+C$ , where C is a constant of integration, then $f(x)$ is equal to :

Option: 1 $\frac{2}{3}(x+2)$
Option: 2 $\frac{1}{3}(x+1)$
Option: 3 $\frac{1}{3}(x+4)$
Option: 4 $\frac{2}{3}(x-4)$

Indefinite integrals for Algebraic functions -

$\frac{\mathrm{d}}{\mathrm{d} x} \frac{\left ( x^{n+1} \right )}{n+1}=x^{n}$ so $\int x^{n}dx=\frac{x^{n+1}}{n+1}$

- wherein

Where $n\neq-1$

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable , $\left ( t\, or\ \theta \right )$

Let t = $\sqrt{2x-1}$

$=>t^{2}={2x-1}$

$=>2t\: dt=2\: dx$

$\int \frac{x+1}{\sqrt{2x-1}}dx=\int \frac{t^{2}+3}{2}dt$

$=\frac{1}{2}\left [ \frac{t^{3}}{3}+3t \right ]$

$=\frac{t}{6}(t^{2}+9)+C$

t = $\sqrt{2x-1}$

$=\frac{\sqrt{2x-1}}{6}(2x+8)+C$

$=\frac{\sqrt{2x-1}}{3}(x+4)+C$

$=>f (x)=\frac{x+4}{3}$

View Full Answer(1)

Posted by

Ritika Jonwal

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

If f(a+b+1-x)=f(x), for all x, where a and b are fixed positive real numbers, then $\frac{1}{a+b}\int_{a}^{b}x(f(x)+f(x+1))dx$ is equal to :
Option: 1 $\int_{a-1}^{b-1}f(x)dx$

Option: 2 $\int_{a+1}^{b+1}f(x+1)dx$

Option: 3 $\int_{a-1}^{b-1}f(x+1)dx$

Option: 4 $\int_{a+1}^{b+1}f(x)dx$

Properties of the Definite Integral (Part 2) - King's Property -

Property 4 (King's Property)

This is one of the most important properties of definite integration.

$\\\mathbf{\int_{a}^{b}f(x)\;dx=\int_{a}^{b}f(a+b-x)\;dx}$

$f(a+b+1-x)=f(x)$

$\text{put x =1+x}$

$f(a+b-x)=f(x+1)$

$I=\frac{1}{a+b}\int_{a}^{b}\left [x(f(x)+f(x+1)) \right ]dx$ ...........1

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b-x)(f(a+b-x)+f(a+b-x+1)) \right ]dx$

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b-x)(f(x+1)+f(x)) \right ]dx$ ..........2

Adding 1 and 2

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b)(f(x+1)+f(x)) \right ]dx$

$I=\int_{a}^{b}\left [(f(x+1)+f(x)) \right ]dx$

$\begin{array}{l}{2 \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}+1) \mathrm{d} \mathrm{x}+\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}} \\ 2I={\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}+1-\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}} \\ {2 \mathrm{I}=2 \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}}\end{array}$

Correct Option (4)

View Full Answer(1)

Posted by

Ritika Jonwal

The area of the region, enclosed by the circle $x^{2}+y^{2}=2$ which is not common to the region bounded by the parabola $y^{2}=x$ and the straight line $y=x,$ is :

Option: 1 $\frac{1}{3}(12\pi -1)$

Option: 2 $\frac{1}{6}(12\pi -1)$

Option: 3 $\frac{1}{3}(6\pi -1)$

Option: 4 $\frac{1}{6}(24\pi -1)$

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves y = f(x), y = g(x) and intersect each other in the interval [a, b]

First find the point of intersection of these curves y = f(x) and y = g(x) , let the point of intersection be x = c

Area of the shaded region

$=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x$

When two curves intersects more than one point

rea bounded by the curves y=f(x), y=g(x) and intersect each other at three points at x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

$\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx$

$\\x^2+y^2=2\Rightarrow r=\sqrt 2\\y^2=x\\y=x\\$

Area between parabola and line is A1

$\\A1=\int_{0}^{1}(y^2-y)dx\\A1=\int_{0}^{1}(\sqrt x-x)dx=1/6\\\text{Required area }=2\pi-1/6$

Correct Option (2)

View Full Answer(1)

Posted by

Ritika Jonwal

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Ranking

Products & Resources

NCERT Solutions

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Animation Courses

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

All Questions

Given : and Then the area (in sq. units) of the region bounded b the curves, and between the lines, is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

Let a function be continuous, and F be defined as : , where Then for the function F, the point is : Option: 1 a point of infection. Option: 2 a point of local maxima. Option: 3 a point of local minima. Option: 4 not a critical point.

Posted by

avinash.dongre

Crack CUET with india's "Best Teachers"

The value of is equal to : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

is equal to Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

JEE Main high-scoring chapters and topics

If where c is a constant of integration, then is equal to : Option: 1 Option: 2 Option: 3 Option: 4

The value of $\int_{0}^{2x}\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx$ is equal to :
Option: 1 $2\pi$

Option: 2 $4\pi$

Option: 3 $2\pi ^{2}$

Option: 4 $\pi ^{2}$

$\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}$ is equal to
Option: 1 $\frac{18}{e^{4}}$
Option: 2 $\frac{27}{e^{2}}$
Option: 3 $3\log ^{3-2}$
Option: 4 $3\log ^{3-2}$

If $\int \frac{\cos \: x\: dx}{\sin ^{3}x(1+\sin ^{6}x)^{2/3}}= f(x)(1+ \sin ^{6}x)^{1/\lambda }+c$ where c is a constant of integration, then $\lambda f\left ( \frac{\pi }{3} \right )$ is equal to :
Option: 1 $-\frac{9}{8}$
Option: 2 $\frac{9}{8}$
Option: 3 $2$
Option: 4 $-2$

If $\int \frac{x+1}{\sqrt{2x-1}}dx=f(x)\sqrt{2x-1}+C$ , where C is a constant of integration, then $f(x)$ is equal to :

Option: 1 $\frac{2}{3}(x+2)$
Option: 2 $\frac{1}{3}(x+1)$
Option: 3 $\frac{1}{3}(x+4)$
Option: 4 $\frac{2}{3}(x-4)$

If f(a+b+1-x)=f(x), for all x, where a and b are fixed positive real numbers, then $\frac{1}{a+b}\int_{a}^{b}x(f(x)+f(x+1))dx$ is equal to :
Option: 1 $\int_{a-1}^{b-1}f(x)dx$

Option: 2 $\int_{a+1}^{b+1}f(x+1)dx$

Option: 3 $\int_{a-1}^{b-1}f(x+1)dx$

Option: 4 $\int_{a+1}^{b+1}f(x)dx$