Inter 2nd Year Maths Exercise 7g Solutions

by

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7g will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7g

Integrals Exercise 7g Solutions

I. Integrate the following functions.

Question 1.
\(\sqrt{4-x^2}\)
Solution:
Let I = \(\int \sqrt{4-x^2}\) dx
= \(\int \sqrt{(2)^2-(x)^2}\) dx [∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)]
∴ I = \(\frac{x}{2} \sqrt{4-x^2}+2 \sin ^{-1} \frac{x}{2}+C\)

Inter 2nd Year Maths Exercise 7g Solutions

Question 2.
\(\sqrt{1-4 x^2}\)
Solution:
Let I = \(\int \sqrt{1-4 \mathrm{x}^2} \mathrm{dx}\)
= ∫ \(\sqrt{4\left(\frac{1}{4}-x^2\right)}\) dx = 2 ∫\(\sqrt{\left(\left(\frac{1}{2}\right)^2-x^2\right)}\) dx
= \(2 \cdot \frac{x}{2} \sqrt{\frac{1}{4}-x^2}+\frac{1}{4} \cdot \frac{2}{2} \sin ^{-1} \frac{x}{1 / 2}+C\)
[∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{2}+C\)]
∴ I = \(\frac{x}{2} \sqrt{1-4 x^2}+\frac{1}{4} \sin ^{-1} 2 x+C\)

Question 3.
\(\sqrt{x^2+4 x+6}\)
Solution:
Inter 2nd Year Maths Exercise 7g Solutions 1

Question 4.
\(\sqrt{x^2+4 x+1}\)
Solution:
Let I = ∫ \(\sqrt{x^2+4 x+1} d x=\int \sqrt{\left(x^2+4 x+4\right)-3}\) dx
= ∫\(\sqrt{x^2+4 x+1-2^2+2^2}\) dx = ∫\(\sqrt{(x+2)^2-(\sqrt{3})^2}\) dx
∴ I = \(\left(\frac{x+2}{2}\right) \sqrt{x^2+4 x+1}-\frac{3}{2} \log (x+2)+\sqrt{x^2+4 x+1}\) + C
[∵ \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|\)]

Inter 2nd Year Maths Exercise 7g Solutions

Question 5.
\(\sqrt{1-4 x-x^2}\)
Solution:
Let I = ∫\(\sqrt{1-4 x-x^2}\) dx = ∫\(\sqrt{-\left(x^2+4 x-1-2^2+2^2\right)}\) dx
= ∫\(-\left[(x+2)^2-(\sqrt{5})^2\right]\) dx
= ∫\(\sqrt{(\sqrt{5})^2-(x+2)^2}\) dx
∴ I = \(\frac{(x+2)}{2} \sqrt{1-4 x-x^2}+\frac{5}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+C\)
[∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)]

Question 6.
\(\sqrt{x^2+4 x-5}\)
Solution:
Let I = ∫\(\sqrt{x^2+4 x-5-2^2}+2^2\) dx
= ∫\(\sqrt{(x+2)^2-5-4}\) dx = ∫\(\sqrt{(x+2)^2-(3)^2}\) dx
∴ I = \(\frac{(x+2)}{2} \sqrt{x^2+4 x-5}-\frac{9}{2} \log \left|(x+2)+\sqrt{x^2+4 x-5}\right|\) + C
[∵ \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|\)]

Inter 2nd Year Maths Exercise 7g Solutions

Question 7.
\(\sqrt{1+3 x-x^2}\)
Solution:
Inter 2nd Year Maths Exercise 7g Solutions 2

Inter 2nd Year Maths Exercise 7g Solutions

Question 8.
\(\sqrt{x^2+3 x}\)
Solution:
Inter 2nd Year Maths Exercise 7g Solutions 3

Inter 2nd Year Maths Exercise 7g Solutions

Question 9.
\(\sqrt{1+\frac{x^2}{9}}\)
Solution:
Inter 2nd Year Maths Exercise 7g Solutions 4

Leave a Comment