Mathematics

# Find the area bounded by the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ and the ordinates $x=0$ and $x=ae$, where $b^{2}=a^{2}(1-e^{2})$ and $e<1$

##### SOLUTION
Required area $A$ is given by
$A=2$ (Area of shaded region in first quadrant)
$\Rightarrow A=2 \displaystyle\int_{0}^{ae}|y|\ dx=2\displaystyle\int_{0}^{ae}y\ dx$                      $[\because y\ge 0\therefore |y|=y]$
$\Rightarrow A=2\dfrac{b}{a}\left[\displaystyle\int_{0}^{ae}\sqrt{a^{2}-x^{2}}dx\right]$
$\Rightarrow A=\dfrac{2b}{a}\left[\dfrac{1}{2}\times \sqrt{a^{2}-x^{2}}+\dfrac{1}{2}a^{2}\sin^{-1}\dfrac{x}{a}\right]_{ae}^{0}$
$\Rightarrow A=\dfrac{2b}{a}\left[\dfrac{ae}{2}\sqrt{a^{2}-a^{2}e^{2}}+\dfrac{1}{2}a^{2}\sin^{-1}e\right]$
$\Rightarrow A=\dfrac{b}{a}\left(a^{2}e\sqrt{1-e^{2}}+a^{2}\sin^{-1}e\right)=ab\left(e\sqrt{1-e^{2}}+\sin^{-1}e\right)$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Hard
A function $f(x)$ which satisfies the relation $f(x) = e^{x} + \int_{0}^{1} e^{x}f(t)dt$, then $f(x)$ is
• A. $(e - 2)e^{x}$
• B. $2e^{x}$
• C. $\dfrac {e^{x}}{2}$
• D. $\dfrac {e^{x}}{2 - e}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The integral $\displaystyle \int { \left( 1+2{ x }^{ 2 }+\frac { 1 }{ x } \right) } { e }^{ { x }^{ 2-\frac { 1 }{ x } } }dx$ is equal to
• A. $(2x-1).e^{x^{2-\dfrac {1}{x}}}+c$
• B. $(2x+1).e^{x^{2-\dfrac {1}{x}}}+c$
• C. $-xe^{x^{2-\dfrac {1}{x}}}+c$
• D. $xe^{x^{2-\dfrac {1}{x}}}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int \frac{1}{[(x-1)^3(x-2)^5]^{\frac{1}{4}}} dx$ is equal to
• A. $\displaystyle \frac{4}{3} \left ( \frac{x+2}{x-1} \right )^{\frac{1}{4}}+C$
• B. $\displaystyle \frac{1}{4} \left ( \frac{x-1}{x+2} \right )^{\frac{1}{4}}+C$
• C. $\displaystyle \frac{1}{3} \left ( \frac{x+2}{x-1} \right )^{\frac{1}{4}}+C$
• D. $\displaystyle \frac{4}{3} \left ( \frac{x-1}{x+2} \right )^{\frac{1}{4}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Passage Hard
Integral of form $\int f\left ( x, \sqrt{ax^{2}+bx+c} \right )dx$ can be evaluated with the help of Euler's substitution in the following manners:
(a) If $a> 0$, we put $\sqrt{ax^{2}+bx+c}=t\pm x\sqrt{a}$ or $ax^{2}+bx+c=t^{2}\pm tx\sqrt{a}+ax^{2}$ i.e. $bx+c=t^{2}\pm 2tx\sqrt{a}$
(b) If $c> 0$, we put $\sqrt{ax^{2}+bx+c}=tx\pm \sqrt c$ or $ax+b=t^{2}x\pm 2tx\sqrt{c}$
(c) If the trinomial $ax^{2}+bx+c$ has $\alpha$, $\beta$ as its real zero's
i.e. $ax^{2}+bx+c=a\left ( x-\alpha \right )\left ( x-\beta \right )$ then we put $\sqrt{ax^{2}+bx+c}=t\left ( x-\alpha \right )$ or $t\left ( x-\beta \right )$
On the basis of above information answer the following question:

Let g(x) =$\displaystyle \int_{0}^{x}f\left ( t \right )dt,$ where f is a function