Samacheer Kalvi 11th Maths Guide Chapter 11 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 11.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 கணங்கள், தொடர்புகள் மற்றும் naசார்புகள் Ex 11.1 Textbook Questions and Answers, Notes.

TN Board 11th Maths Solutions Chapter 11 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 11.1

கீழ்க்காண்பனவற்றைத் தொகையிடுக :

Question 1.
(i) x¹¹

(ii) \(\frac{1}{x^{7}}\)

(iii) \(\sqrt[3]{x^{4}}\)

(iv) \(\left(x^{5}\right)^{\frac{1}{8}}\)

(i) x¹¹
தீர்வு :
∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + c
∴ ∫ x¹¹ dx = \(\frac{x^{11+1}}{11+1}\) + c
= \(\frac{x^{12}}{12}\) + c

(ii) ∫ \(\frac{1}{x^{7}}\) dx = ∫x^-7 dx
= \(\frac{x^{-7+1}}{-7+1}\)
= \(\frac{x^{-6}}{-6}\) + c
= \(-\frac{1}{6 x^{6}}\) + c

(iii) ∫ \(\sqrt[3]{x^{4}}\) dx = ∫ (x⁴)^{\(\frac{1}{3}\)} dx
= x^{\(\frac{4}{3}\)} dx

= \(\frac{x^{\frac{4}{3}}+1}{\frac{4}{3}+1}\) + c

= \(\frac{x^{\frac{7}{3}}}{\frac{7}{3}}\) + c

= \(\frac{3}{7} x^{\frac{7}{3}}\) + c

(iv) ∫ \(\left(x^{5}\right)^{\frac{1}{8}}\) dx
= ∫ x^{\(\frac{5}{8}\)} dx

= \(\frac{x^{\frac{5}{8}+1}}{\frac{5}{8}+1}\)+ c

= \(\frac{x^{\frac{13}{8}}}{\frac{13}{8}}\) + c

= \(\frac{8}{13} x^{\frac{13}{8}}\) + c

Question 2.
(i) \(\frac{1}{\sin ^{2} x}\)

(ii) \(\frac{\tan x}{\cos x}\)

(iii) \(\frac{\cos x}{\sin ^{2} x}\)

(iv) \(\frac{1}{\cos ^{2} x}\)
தீர்வு :
\(\frac{1}{\sin ^{2} x}\)

∫ \(\frac{1}{\sin ^{2} x}\) dx = \(\frac{1}{\sin ^{2} x}\) cosec² dx = – cot x + c

(ii) \(\frac{\tan x}{\cos x}\)

∫ \(\frac{\tan x}{\cos x}\) dx = ∫ tan x × \(\frac{1}{\cos x}\) dx
= ∫ tan x sec x dx = sec x + c

(iii) \(\frac{\cos x}{\sin ^{2} x}\)

∫ \(\frac{\cos x}{\sin ^{2} x}\) dx = ∫ \(\frac{\cos x}{\sin x} \times \frac{1}{\sin x}\) dx
= ∫ cot x cosec x dx = – cosec x + c

(iv) \(\frac{1}{\cos ^{2} x}\)
∫ \(\frac{1}{\cos ^{2} x}\) dx = ∫ sec² dx = tan x + c

Question 3.
(i) 12³

(ii) \(\frac{x^{24}}{x^{25}}\)

(iii) e^x
தீர்வு :

(i) 12³ = ∫ 12³ dx = 12³ ∫ dx = 12³ + c

(ii) \(\frac{x^{24}}{x^{25}}\)

= ∫ \(\frac{x^{24}}{x^{25}}\) dx = ∫ \(\frac{1}{x}\) dx
= log |x| + c

(iii) e^x
= ∫ e^x dx = e^x + c

Question 4.
(i) (1 + x²)^-1

(ii) (1 – x²)^{\(-\frac{1}{2}\)}
தீர்வு :
(i) (1 + x²)^-1

= ∫ (1 + x²)^-1 dx

= ∫ \(\frac{1}{1+x^{2}}\) dx = tan^-1 x + c

(ii) ∫ (1 – x²)^{\(-\frac{1}{2}\)} dx

= ∫ \(\frac{1}{\left(1-x^{2}\right)^{\frac{1}{2}}}\) dx

= ∫ \(\frac{d x}{\sqrt{1-x^{2}}}\) + cOS

= sin^-1 + c