Question
Using integration, find the area of the region: $\left\{(\text{x},\text{y}) : |\text{x}-1|<\text{y}<\sqrt{5-\text{x}^{2}}\right\}$.
✓
Answer
$|\text{x}-1|<\text{y}<\sqrt{5-\text{x}^{2}}$$|\text{x}-1|=\sqrt{5-\text{x}^{2}}$
$\text{x}=2, -1$
$\text{A}=\int\limits_{-1}^{2} \Big(\sqrt{5-\text{x}^{2}}-|\text{x}-1|\Big)\text{ dx}$
$=\int\limits_{-1}^{2} \sqrt{5-\text{x}^{2}}+\int\limits_{-1}^{2}|\text{x}-1|\text{ dx}+\int\limits_{-1}^{2}(1-\text{x})\text{ dx}$
$=\Big[\frac{\text{x}}{2} \sqrt{5-\text{x}^{2}}+\frac{5}{2}\sin^{-1}\Big(\frac{\text{x}}{5}\Big)\Big]^{2}_{-1}+\Big[\frac{\text{x}^{2}}{2}-\text{x}\Big]^{1}_{-1}+\Big[\text{x}-\frac{\text{x}^{2}}{2}\Big]^{2}_{1}$
$=\frac{5}{2}\sin^{-1}\Big(\frac{\text{2}}{\sqrt5}\Big)+\sin^{-1}\Big(\frac{\text{1}}{\sqrt5}\Big)+\frac{1}{2}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
→$\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
Show that the following system of linear equations is consistent and also find solutions:x - y + z = 3
2x + y - z = 2
-x - 2y + 2z = 1
2x + y - z = 2
-x - 2y + 2z = 1
A bag A contains $5$ white and $6$ black balls. Another bag B contains $4$ white and $3$ black balls. A ball is transferred from bag A to the bag B and then a ball is taken out of the second bag. Find the probability of this ball being black.
→Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^4\text{x dx}$
→A(1, 0, 4), B(0, -11, 13), C(2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.
Differentiate the following functions with respect to x:
$(\log\text{x})^\text{x}$
→$(\log\text{x})^\text{x}$
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\big(2\log\cos\text{x}-\log\sin2\text{x}\big)\text{dx}$
→$\int\frac{\text{x}^2+5\text{x}+2}{\text{x}+2}\text{dx}$
→Using integration, find the area of the following region:
$\Big\{(\text{x},\text{y}):\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}\leq1\leq\frac{\text{x}}{3}+\frac{\text{y}}{2}\Big\}$
→$\Big\{(\text{x},\text{y}):\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}\leq1\leq\frac{\text{x}}{3}+\frac{\text{y}}{2}\Big\}$
The vertices A, B, C of triangle ABC have respectively position vector $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ with respect to given origin O. Show that the point D where the bisector of $\angle{\text{A}}$ meets BC has position Vector $\vec{\text{d}}=\frac{\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\beta+\gamma}$, where $\beta=\big|\vec{\text{c}}-\vec{\text{a}}\big|$ and, $\gamma=\big|\vec{\text{a}}-\vec{\text{b}}\big|$.
→