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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 - $\int_{1/2}^{\sqrt3/2}\left ( x-\frac{1}{2} \right )^2dx$

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

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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).

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

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

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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.

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

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

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

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$\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}$ 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

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

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

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

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

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

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

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

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

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

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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

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

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

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

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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)

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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)

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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 = $\sqrt{2x-1}$

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

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

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

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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)

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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)

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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

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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.

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

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avinash.dongre

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

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