Let I = ……….(i) Let ……….(ii) Comparing coefficients of A + C = 1 ……….(iii) Comparing coefficients of 3A + B + 2C = 1 ……….(iv) Comparing constants 2A + 2B + C = 1 ……….(v) On solving...
Let I = = = = = = =
Let I = Putting I = = = = I = ……….(i) I = = I = I = ……….(ii) For evaluating , putting = = = = ……….(iii) Putting this value in eq. (ii), I = I = = I...
We know that I = = I = = ……….(i) Putting = I = = I = [Applying product rule] I = I = I = Putting I = I = I...
Let I = = I = = I = Putting I = = = I = = I = =
Let I = = ………..(i) Putting From eq. (i), I = = if And I = = = if I = if And I = if
Let I = = = = I = Putting From eq. (i), I = = I = =
Let I = = = ……….(i) Putting From eq. (i), I = = =
Let I = ……….(i) I = = I = = I = +c
Let I = ……….(i) Putting From eq. (i), I = = ……….(ii) I = = I = = I = =
Let I = = ……….(i) Putting From eq. (i), I = = =
Let I = ……….(i) I = = I = I = I = I = I =
Let I = ……….(i) I = = I = I = I = = I = =
Let I = ……….(i) Putting From eq. (i), I = = =
Let I = = = = I = =
Let I = = = = = = = = =
Let I = ……….(i) Let ……….(ii) Comparing coefficients of A + B = 0 ……….(iii) Comparing coefficients of B + C = 5 ……….(iv) Comparing constants 9A + C = 0 ……….(v) On solving eq. (iii), (iv) and...
Let I = Putting I = = = = = = = = = =
Let I = = = = Putting I = = =
Let I = ……….(i) Putting From eq. (i), I = = = = = =
Solution: Given equations are $y=\cos x \ldots(1)$ And, $y=\sin x \ldots$ $\text { Required area }=\text { Area (ABLA) }+\text { area (OBLO) }$ $\begin{array}{l} =\int_{1}^{1} x d...
Solution: Given equations are $x^{2}+y^{2}=16 \dots \dots (1)$ $y^{2}=6x \dots \dots (2)$ The area bounded by the circle and parabola $=2[\text { Area }(\mathrm{OADO})+\text { Area...
Solution: $\text { Required area }=\int_{-1}^{1} y d x$ $\begin{array}{l} =\int_{-1}^{1} x|x| d x \\ =\int_{-1}^{0} x^{2} d x+\int_{0}^{1} x^{2} d x \end{array}$ $\begin{array}{l}...
Solution: $\begin{array}{l} \text { Required area }=\int_{-2}^{1} y d x \\ =\int_{-2}^{1} x^{3} d x \\ =\left[\frac{x^{4}}{4}\right]_{-2}^{1} \end{array}$ $\begin{array}{l}...
Solution: Eq. of parabola is $y^{2}=4 x$......(i) Eq. of circle is $4 x^{2}+4 y^{2}=9 \ldots \ldots$ (ii) From the diagram, the points of intersection of parabola (i) and circle (ii) are...
Solution: Given eqs. of lines are $\begin{array}{l} 2 x+y=4 \ldots(1) \\ 3 x-2 y=6 \ldots(2) \end{array}$ And, $x-3 y+5=0 \ldots$ (3) Area of the region bounded by the lines is the area of...
Solution: Vertices of $\triangle A B C$ are $A(2,0), B(4,5)$, and $C(6,3)$. Eq. of line segment $A B$ is $\begin{array}{l} y-0=\frac{5-0}{4-2}(x-2) \\ 2 y=5 x-10 \\ y=\frac{5}{2}(x-2) \end{array}$...
Solution: Area bounded by the curves, $\left\{(x, y): y \geq x^{2}\right.$ and $\left.y=|x|\right\}$, is represented by the shaded region as It can be observed that the required area is symmetrical...
Solution: Area bounded by the curve, $|x|+|y|=1$, is represented by the shaded region $\mathrm{ADCB}$ as The curve intersects the axes at points $A(0,1), B(1,0), C(0,-1)$, and $D(-1,0)$. It can be...
Solution: Area of the region enclosed by the parabola, $x^{2}=y$, the line, $y=x+2$, and $x$-axis is represented bv the shaded reaion $\mathrm{OABCO}$ as Point of intersection of the parabola,...
Solution: Area of the smaller region bounded by the ellipse, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the line, $\frac{x}{a}+\frac{y}{b}=1$ is represented by the shaded region BCAB as...
Solution: Area of the smaller region bounded by the ellipse, $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and the line, $\frac{x}{3}+\frac{y}{2}=1$ is represented by the shaded region BCAB as $\therefore$...
Solution: Area enclosed between the parabola, $4 y=3 x^{2}$, and the line, $2 y=3 x+12$, is represented by the shaded area OBAO as Points of intersection of the given curves are $A(-2,3)$ and...
Solution: Area enclosed between the parabola, $y^{2}=4 a x$, and the line, $y=m x$, is represented by the shaded area $\mathrm{OABO}$ as Points of intersection of both the curves are $(0,0)$ and...
Solution: Graph of $y=\sin x$ can be drawn as $\therefore$ Required area $=$ Area OABO $+$ Area BCDB $\begin{array}{l} =\int_{0}^{\pi} \sin x d x+\left|\int_{k}^{2 \pi} \sin x d x\right| \\ =[-\cos...
Solution: Given eq. is $y=|x+3|$ Corresponding values of $x$ and $y$ are given in the following table. $$\begin{tabular}{|l|l|l|l|l|l|l|l|} \hline$x$ & $-6$ & $-5$ & $-4$ & $-3$ & $-2$ & $-1$ & 0 \\...
Solution: The area in the first quadrant is bounded by $y=4 x^{2}, x=0, y=1$, and $y=4$ is represented by the shaded area ABCDA as $\begin{aligned} \therefore \operatorname{Area} A B C D...
Solution: The area required is represented by the shaded area OBAO as The point of intersection of the curves $y=x$ and $y=x^{2}$ is A(1,1) Draw AC perpendicular to x-axis $\therefore$ Area...
Solution: (i)The area required is represented by the shaded area ADCBA as $\begin{aligned} \text { Area } \mathrm{ADCBA} &=\int^{2} y d x \\ &=\int^{2} x^{2} d x \\...
Solution: The area lying between the curve, $y^{2}=4 x$ and $y=2 x$, is represented by the shaded area OBAO as Points of intersection of these curves are $O(0,0)$ and $A(1,2)$. Draw AC perpendicular...
Solution: The smaller area enclosed by the circle, $x^{2}+y^{2}=4$, and the line, $x+y=2$, is represented by the shaded area ACBA as It is observed that, $\text { Area } A C B A=\text { Area } O A C...
Solution: The eqs. of the sides of the triangle are $y=2 x+1, y=3 x+1$, and $x=4$. On solving these eqs., we get the vertices of triangle as $\mathrm{A}(0,1), \mathrm{B}(4,13)$, and $\mathrm{C}$...
Solution: $\mathrm{BL}$ and $\mathrm{CM}$ are drawn perpendicular to $x$-axis. It is observed in the following figure that, Area $(\triangle \mathrm{ACB})=$ Area (ALBA) $+$ Area (BLMCB) - Area...
Solution: The area of the region bounded by the curves $y=x^{2}+2$, $y=x$, $x=0$ and $x=3$ is represented by the shaded area OCBAO as Therefore, Area of OCBAO = Area of ODBAO - Area of ODCO...
Solution: The area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is reprsented by the shaded area as On solving the equations, $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$, we get the...
Solution: Required area is represented by the shaded area OBCDO On solving the given equation of circle, $4 x^{2}+4 y^{2}=9$, and parabola, $x^{2}=4 y$, we get $\mathrm{B}\left(\sqrt{2},...
Solution: Area bounded by the curve, $y^{2}=4 x, y$-axis, and $y=3$ is represented as $\begin{aligned} \therefore \text { Area } \mathrm{OAB} &=\int_{0}^{3} x d y \\ &=\int_{0}^{3}...
Solution: The area bounded by the circle and the lines, $x=0$ and $x=2$, in the first quadrant is represented as $\begin{aligned} \therefore \text { Area } \mathrm{OAB} &=\int_{0}^{2} y d x \\...
Solution: The region bounded by the parabola, $y^{2}=4 x$, and the line, $x=3$, is the area $\mathrm{OACO}$. Area of OACO is symmetrical about $x$-axis. $\therefore$ Area of $O A C O=2$ (Area of...
Solution: The area bounded by the curve, $x^{2}=4 y$, and line, $x=4 y-2$, is represented by the shaded area OBAO. Let's say $A$ and $B$ be the points of intersection of the line and parabola....
Solution: The area bounded by the parabola, $x^{2}=y$, and the line, $y=|x|$, can be represented as The given area is symmetrical about $y$-axis. $\therefore$ Area $\mathrm{OACO}=$ Area...
Solution: The line, $x=a$, divides the area bounded by the parabola and $x=4$ into two equal parts. $\therefore$ Area $O A D=$ Area $A B C D$ It is observed that the given area is symmetrical about...
Solution: The area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cutt off by the line $x=\frac{a}{\sqrt{2}}$ is the area of the ABCDA It can be observed that the area $A B C D$ is...
Solution: The area of the region bounded by the circle, $x^{2}+y^{2}=4, x=\sqrt{3} y$, and the $x$-axis is the area OAB. The point of intersection of the line and the circle in the first quadrant is...
Solution: The given eq. of the ellipse can be represented as $\begin{array}{l} \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 \\ \Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}} \end{array}$ It can be observed that the...
Solution: The given eq. of the ellipse, $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1$ can be reprsented by It is observed that the ellipse is symmetrical about x-axis and y-axis. $\therefore$ Area...
Solution: The area of the region bounded by $x^{2}=4y$, $y=2$, $y=4$ and the y-axis is the area of ABCD
Solution: Area of the region bounded by the curve $y^{2}=9x$, $x=2$, $x=4$, \and the x-axis is the area of ABCD
Solution:
The equation of curves are y2 = 2x and x2 + y2 = 4x Solving the equations, \[\begin{array}{*{35}{l}} {{x}^{2~}}-\text{ }4x\text{ }+\text{ }{{y}^{2~}}=\text{ }0 \\ {{x}^{2~}}-\text{ }4x\text{...
The curves are y = √x and line x = 2y + 3 Solving y = √x and x = 2y + 3, we get
The curve y = –x2 or x2 = –y and the line x + y + 2 = 0 Solving the two equation, we get \[x\text{ }-\text{ }{{x}^{2}}~+\text{ }2\text{ }=\text{ }0\] \[{{x}^{2}}~\text{ }-x\text{ }\text{ }-2\text{...
The equations of curve y = √x and line y = x Solving the equations y = √x ⇒ y2 = x and y = x, we get \[\begin{array}{*{35}{l}} {{x}^{2}}~=\text{ }x \\ {{x}^{2}}~-\text{ }x\text{ }=\text{ }0 \\...
\[\begin{array}{*{35}{l}} y~=\text{ }\surd \left( {{a}^{2}}~-~{{x}^{2}} \right)\text{ }\Rightarrow ~{{y}^{2}}~=~{{a}^{2}}~-~{{x}^{2}} \\ {{x}^{2~}}+~{{y}^{2}}~=~{{a}^{2}} \\ \end{array}\] which is...
The curve is \[\begin{array}{*{35}{l}} y~=\text{ }\surd \left( x~-\text{ }1 \right) \\ \Rightarrow \text{ }{{y}^{2}}~=\text{ }x\text{ }-\text{ }1 \\ \end{array}\] Plotting the curve and finding...
Given, \[2y~=\text{ }5x~+\text{ }7,~x-axis,~x~=\text{ }2\text{ }and~x~=\text{ }8\] graphg for: \[2y~=\text{ }5x~+\text{ }7\text{ }\Rightarrow \text{ }y\text{ }=\text{ }\left( 5x\text{ }+\text{ }7...
Given, {(x, 0) : \[\begin{array}{*{35}{l}} y~=\text{ }\surd \left( 4\text{ }~-{{x}^{2}} \right)\ \\ So,\text{ }{{y}^{2}}~=\text{ }4\text{ }-\text{ }{{x}^{2}} \\ {{x}^{2}}~+\text{...
Given, equation of parabola x2 = y and line y = x + 2 Solving the above equations, we get \[\begin{array}{*{35}{l}} {{x}^{2}}~=\text{ }x\text{ }+\text{ }2 \\ {{x}^{2}}~-\text{ }x\text{ }-\text{...
The curves are y2 = 9x and y = x Solving the above equations, we have \[\begin{array}{*{35}{l}} {{x}^{2}}~=\text{ }9x\text{ }\Rightarrow \text{ }{{x}^{2}}~\text{ }-9x\text{ }=\text{ }0 \\ x\left(...
The curves are y2 = 4x … (i) and x2 = 4y … (ii) On solving the equations, we get From (ii), y = x2/4 Putting value of y in (i), we have \[\begin{array}{*{35}{l}} {{\left( {{x}^{2}}/4...
The curves are y = x3, y = x + 6 and x = 0 On solving y = x3 and y = x + 6, we have \[\begin{array}{*{35}{l}} {{x}^{3}}~=\text{ }x\text{ }+\text{ }6 \\ {{x}^{3}}~\text{ }-x\text{ }-\text{ }6\text{...
equations of parabolas are \[\begin{array}{*{35}{l}} {{y}^{2}}~=\text{ }2px~\ldots .\text{ }\left( i \right) \\ ~{{x}^{2}}~=\text{ }2py~\ldots .\text{ }\left( ii \right) \\ \end{array}\] Now, from...
Given curves are y2 = 9x and y = 3x solving the two equations we have \[\begin{array}{*{35}{l}} {{\left( 3x \right)}^{2}}~=\text{ }9x \\ 9{{x}^{2}}~=\text{ }9x \\ 9{{x}^{2}}~\text{ }-9x\text{...