Question

evaluate each of the following integrals.
chp- Definite integrals​

Answer 1

AnsWer :

π / 2.

QuestioN :

$\tt1. \int\limits_{0}^{1 /2}\dfrac{dx}{ \sqrt{4 - {x}^{2} } }$

SolutioN :

$\tt \longmapsto\int\limits_{0}^{1 /2}\dfrac{dx}{ \sqrt{4 - {x}^{2} } }$

$\tt \longmapsto\int\limits_{0}^{1 /2}\dfrac{dx}{ \sqrt{ {(2)}^{2} - {x}^{2} } }$

★ Convert into :

• a² - b² = ( a + b )( a - b )

We have, Formula.

• Sin^-1 x / 2.
• Where as,
• a = 2.

$\tt \longmapsto{Sin}^{-1} \frac{x}{2}\Bigg|_{0}^{2}$

Now, Putting the value of 2 to 0.

$\tt \longmapsto{Sin}^{-1} \frac{2}{2} -{Sin}^{-1} \frac{0}{2}$

$\tt \longmapsto{Sin}^{-1} 1 -{Sin}^{-1} 0$

$\tt \longmapsto\dfrac{\pi}{2}-0.$

$\tt \longmapsto\dfrac{\pi}{2}.$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\boxed{\sf{\int _0^{\tfrac{1}{2}}\dfrac{1}{\sqrt{4-x^2}}dx}}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int _0^{\tfrac{1}{2}}\dfrac{1}{\sqrt{4-x^2}}dx=\arcsin \left(\dfrac{1}{4}\right)\quad \left(\mathrm{Decimal:\quad }\:0.25268\dots \right)}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Apply Trig Substitution: } x=2 \sin (u)$

$=\int _0^{\arcsin \left(\tfrac{1}{4}\right)}1du$

$\mathrm{Integral\:of\:a\:constant}:\quad \int adx=ax$

$=\left[1\cdot \:u\right]^{\arcsin \left(\tfrac{1}{4}\right)}_0$

$=\left[u\right]^{\arcsin \left(\tfrac{1}{4}\right)}_0$

$\text { Compute the boundaries: }[u]_{0}^{\arcsin \left(\tfrac{1}{4}\right)}=\arcsin \left(\dfrac{1}{4}\right)$

$\boxed{\sf{=arcsin \left(\dfrac{1}{4}\right)}}$