Area bounded by the curve xy = c and the x-axis between x = 1 and x =

1.	Area bounded by the curve xy = c and the x-axis between x = 1 and x = 4, is:1. c(log 3) sq. units2. 2c(log 3) sq. units3. 2c(log 2) sq. units4. 2c(log 5) sq. units
Answer» Correct Answer - Option 3 : 2c(log 2) sq. units Concept: The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \left\| \int_{a}^{b}f(x)\ dx \right\|\), for curves which are entirely on the same side of the x-axis in the given range. Definite integral: If ∫ f(x) dx = g(x) + C, then \(\rm \int_{a}^{b}f(x)\ dx=[g(x)]_{a}^{b}\) = g(b) - g(a). \(\rm \int{\frac{1}{x}dx}\) = log x + C. Calculation: The given equation is of the curve is xy = c, which can also be written as y = f(x) = \(\rm \frac{c}{x}\). Using definite integrals, the area under the curve from x = 1 to x = 4 and the x-axis, will be given as: A = \(\rm \left\| \int_{1}^{4}{\frac{c}{x}}\ dx \right\|\) Using \(\rm \int{\frac{1}{x}dx}\) = log x + C, we get: ⇒ A = \(\rm c\left[ \log x \right]_{1}^{4}\) ⇒ A = c(log 4 - log 1) Using log 1 = 0 and log 4 = 2 log 2, we get: ⇒ A = 2c(log 2) sq. units

Area bounded by the curve xy = c and the x-axis between x = 1 and x = 4, is:1. c(log 3) sq. units2. 2c(log 3) sq. units3. 2c(log 2) sq. units4. 2c(log 5) sq. units

Answer» Correct Answer - Option 3 : 2c(log 2) sq. units

Concept:

The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \left| \int_{a}^{b}f(x)\ dx \right|\), for curves which are entirely on the same side of the x-axis in the given range.
Definite integral: If ∫ f(x) dx = g(x) + C, then \(\rm \int_{a}^{b}f(x)\ dx=[g(x)]_{a}^{b}\) = g(b) - g(a).
\(\rm \int{\frac{1}{x}dx}\) = log x + C.

Calculation:

The given equation is of the curve is xy = c, which can also be written as y = f(x) = \(\rm \frac{c}{x}\).

Using definite integrals, the area under the curve from x = 1 to x = 4 and the x-axis, will be given as:

A = \(\rm \left| \int_{1}^{4}{\frac{c}{x}}\ dx \right|\)

Using \(\rm \int{\frac{1}{x}dx}\) = log x + C, we get:

⇒ A = \(\rm c\left[ \log x \right]_{1}^{4}\)

⇒ A = c(log 4 - log 1)

Using log 1 = 0 and log 4 = 2 log 2, we get:

⇒ A = 2c(log 2) sq. units