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
577.5k+ 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} $ .
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