Properties of Definite Integrals MCQ Quiz in తెలుగు - Objective Question with Answer for Properties of Definite Integrals - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Apr 29, 2025
Latest Properties of Definite Integrals MCQ Objective Questions
Properties of Definite Integrals Question 1:
\(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) యొక్క విలువను కనుగొనండి
Answer (Detailed Solution Below)
Properties of Definite Integrals Question 1 Detailed Solution
కాన్సెప్ట్:
\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)
సాధన:
I పరిగణించండి= \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) ----(i)
⇒ I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)
⇒ I = -\(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)---(ii)
(i) మరియు (ii) జోడించండి, మేము పొందుతాము
⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}+ \sqrt{sin^8x}}{\sqrt{\cos^8 x}+ \sqrt{\sin^8 x}}dx\)
⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)
⇒ 2I = \(\rm[x]^\frac{\pi}{2}_0\)
⇒ I = \(\dfrac{\pi}{4}\)
Top Properties of Definite Integrals MCQ Objective Questions
\(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) యొక్క విలువను కనుగొనండి
Answer (Detailed Solution Below)
Properties of Definite Integrals Question 2 Detailed Solution
Download Solution PDFకాన్సెప్ట్:
\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)
సాధన:
I పరిగణించండి= \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) ----(i)
⇒ I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)
⇒ I = -\(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)---(ii)
(i) మరియు (ii) జోడించండి, మేము పొందుతాము
⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}+ \sqrt{sin^8x}}{\sqrt{\cos^8 x}+ \sqrt{\sin^8 x}}dx\)
⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)
⇒ 2I = \(\rm[x]^\frac{\pi}{2}_0\)
⇒ I = \(\dfrac{\pi}{4}\)
Properties of Definite Integrals Question 3:
\(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) యొక్క విలువను కనుగొనండి
Answer (Detailed Solution Below)
Properties of Definite Integrals Question 3 Detailed Solution
కాన్సెప్ట్:
\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)
సాధన:
I పరిగణించండి= \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\) ----(i)
⇒ I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)
⇒ I = -\(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)---(ii)
(i) మరియు (ii) జోడించండి, మేము పొందుతాము
⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}+ \sqrt{sin^8x}}{\sqrt{\cos^8 x}+ \sqrt{\sin^8 x}}dx\)
⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)
⇒ 2I = \(\rm[x]^\frac{\pi}{2}_0\)
⇒ I = \(\dfrac{\pi}{4}\)