Area bounded by the curve $ xy=c $ and the x-axis between $ x=1 $ and $ x=4 $ , is:
A. $ c\log 3 $ sq. units
B. $ 2\log c $ sq. units
C. $ 2c\log 2 $ sq. units
D. $ 2c\log 5 $ sq. units
Answer
Verified
450.3k+ views
Hint: The area under the function $ y=f(x) $ from $ x=a $ to $ x=b $ and the x-axis is given by the definite integral $ \left| \int_{a}^{b}{f}(x)\ dx \right| $ , for curves which are entirely on the same side of the x-axis in the given range.
If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
Definite integral: If $ \int{f}(x)dx=g(x)+C $ , then $ \int_{a}^{b}{f}(x)\ dx=[g(x)]_{a}^{b}=g(b)-g(a) $ .
Complete step-by-step answer:
The given equation of the curve is $ xy=c $ which can also be written as $ y=f(x)=\dfrac{c}{x} $ .
Using definite integrals, the area under the curve from $ x=1 $ to $ x=4 $ and the x-axis, will be given as:
$ A=\left| \int_{1}^{4}{\dfrac{c}{x}}\ dx \right| $
Using $ \int{\dfrac{1}{x}dx}=\log x+C $ , we get:
⇒ $ A=c\left[ \log x \right]_{1}^{4} $
⇒ $ A=c(\log 4-\log 1) $
Using $ \log 1=0 $ and $ \log 4=\log {{2}^{2}}=2\log 2 $ , we get:
⇒ $ A=2c\log 2 $ sq. units
The correct answer is C. $ 2c\log 2 $ sq. units.
Note: The graph of $ xy=c $ is a rectangular hyperbola.
In order to calculate the area of a curve from $ y=a $ to $ y=b $ and the y-axis, we will make use of $ \left| \int_{a}^{b}{f}(y)\ dy \right| $ .
The length of a curve $ y=f(x) $ from $ x=a $ to $ x=b $ is given by $ L=\int_{a}^{b}{\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}dx} $ .
If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
Definite integral: If $ \int{f}(x)dx=g(x)+C $ , then $ \int_{a}^{b}{f}(x)\ dx=[g(x)]_{a}^{b}=g(b)-g(a) $ .
Complete step-by-step answer:
The given equation of the curve is $ xy=c $ which can also be written as $ y=f(x)=\dfrac{c}{x} $ .
Using definite integrals, the area under the curve from $ x=1 $ to $ x=4 $ and the x-axis, will be given as:
$ A=\left| \int_{1}^{4}{\dfrac{c}{x}}\ dx \right| $
Using $ \int{\dfrac{1}{x}dx}=\log x+C $ , we get:
⇒ $ A=c\left[ \log x \right]_{1}^{4} $
⇒ $ A=c(\log 4-\log 1) $
Using $ \log 1=0 $ and $ \log 4=\log {{2}^{2}}=2\log 2 $ , we get:
⇒ $ A=2c\log 2 $ sq. units
The correct answer is C. $ 2c\log 2 $ sq. units.
Note: The graph of $ xy=c $ is a rectangular hyperbola.
In order to calculate the area of a curve from $ y=a $ to $ y=b $ and the y-axis, we will make use of $ \left| \int_{a}^{b}{f}(y)\ dy \right| $ .
The length of a curve $ y=f(x) $ from $ x=a $ to $ x=b $ is given by $ L=\int_{a}^{b}{\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}dx} $ .
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Trending doubts
10 examples of friction in our daily life
What problem did Carter face when he reached the mummy class 11 english CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
State and prove Bernoullis theorem class 11 physics CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
Petromyzon belongs to class A Osteichthyes B Chondrichthyes class 11 biology CBSE