1. Tardigrade
  2. Question
  3. Mathematics
  4. The area of the region bounded by the curves x = y2 - 2 and x = y is

Q. The area of the region bounded by the curves $ x = y^2 - 2 $ and $x = y$ is

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Solution:

We have,
$x=y^{2}-2$ and $x=y$
Graph of $x=y^{2}-2$ and $x=y$ are
image
Solving, $x=y^{2}-2$ and $x=y$, we $\operatorname{get}(-1,-1)$ and $(2,2)$
Area of shaded region is
$\int_\limits {-1}^{2} y d y-\int_\limits {-1}^{2}\left(y^{2}-2\right) d y$
$ \Rightarrow \left[\frac{y^{2}}{2}-\frac{y^{3}}{3}+2 y\right]_{-1}^{2} =\left(\frac{4}{2}-\frac{8}{3}+4\right)-\left(\frac{1}{2}+\frac{1}{3}-2\right) $
$=\frac{10}{3}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2} $

Questions from TS EAMCET 2017

1. The mean deviation from the mean $10$ of the data $6,7,10,12,13, \alpha , 12,16$ is

2. Match the following.
List I List II
(I) $\int^{1}_{-1} x |x |dx $ (a) $\frac{\pi}{2}$
(II) $\int^{\frac{\pi}{2}}_{0} \left( 1 + \log\left(\frac{4+3 \sin x }{4+3 \cos x}\right)\right)dx $ (b) $\int^{\frac{\pi}{2}}_{0} f\left(x\right)dx $
(III) $\int^{a}_{0} f(x) dx$ (c) $\int^a_0 [ f(x) + f(-x)] dx $
(IV) $\int^a_{-a} f(x) dx $ (d) 0
(e) $\int^a_0 f(a -x)dx$

3. If $f$ is differentiable, $f(x+y)=f(x) f(y)$ for all $x, y \in I R, f(3)=3, f^{\prime}(0)=11$, then $f^{\prime}(3)=$

4. The probability distribution of a random variable X is given below.
x = k 0 1 2 3 4
P(X = k) 0.1 0.4 0.3 0.2 0

The variance of X is

5. If $A = \begin{bmatrix}1&0&1\\ 0&2&0\\ 1&-1&4\end{bmatrix}, A = B + C , B = B^{T} $ and $C = - C^{T} $, then $ C = $

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