Question

Let $f(x)$: $R^+$ > $R^+$ is an invertible function such that $f'(x) > 0$ and $f''(x) > 0 \forall x \in [1, 5]$ If $f(1) = 1$ and $f(5) = 5$ and area under the curve $y = f(x)$ on x-axis from $x = 1$ to $x = 5$ is 8 sq. units, then area bounded by $y = f^{-1}(x)$ on x-axis from $x = 1$ to $x = 5$ is

Answer 1

16 sq. units

Answer 2

The problem asks for the area bounded by the curve $y = f^{-1}(x)$ on the x-axis from $x = 1$ to $x = 5$. This can be expressed as the definite integral $\int_{1}^{5} f^{-1}(x) dx$.

We are given the following information:

1. $f: R^+ \to R^+$ is an invertible function.
2. $f'(x) > 0$ and $f''(x) > 0$ for all $x \in [1, 5]$. ($f'(x) > 0$ implies $f(x)$ is strictly increasing, hence invertible. $f''(x) > 0$ implies $f(x)$ is convex.)
3. $f(1) = 1$ and $f(5) = 5$.
4. The area under the curve $y = f(x)$ on the x-axis from $x = 1$ to $x = 5$ is 8 sq. units. This means $\int_{1}^{5} f(x) dx = 8$.

To find the area under the inverse function, we use the property of definite integrals involving inverse functions:

For an invertible function $f(x)$ with $f(a) = c$ and $f(b) = d$, the following relationship holds:

$\int_{a}^{b} f(x) dx + \int_{c}^{d} f^{-1}(y) dy = b f(b) - a f(a)$

In this problem, we have:

• $a = 1$
• $b = 5$
• $f(a) = f(1) = 1$ (so $c=1$)
• $f(b) = f(5) = 5$ (so $d=5$)

Substitute these values into the formula:

$\int_{1}^{5} f(x) dx + \int_{1}^{5} f^{-1}(y) dy = 5 \times f(5) - 1 \times f(1)$

We are given $\int_{1}^{5} f(x) dx = 8$. We know $f(1) = 1$ and $f(5) = 5$. Let the required area be $A_{inv} = \int_{1}^{5} f^{-1}(x) dx$. (Note: The variable of integration can be $x$ or $y$; it does not change the value of the definite integral).

Substitute the known values into the equation:

$8 + A_{inv} = 5 \times 5 - 1 \times 1$ $8 + A_{inv} = 25 - 1$ $8 + A_{inv} = 24$

Now, solve for $A_{inv}$:

$A_{inv} = 24 - 8$ $A_{inv} = 16$

Thus, the area bounded by $y = f^{-1}(x)$ on the x-axis from $x = 1$ to $x = 5$ is 16 sq. units.

The conditions $f'(x) > 0$ and $f''(x) > 0$ ensure that the function is strictly increasing and convex. Since $f(1)=1$ and $f(5)=5$, and $f''(x) > 0$, the function $f(x)$ lies below the line $y=x$ for $x \in (1,5)$. This is consistent with $\int_{1}^{5} f(x) dx = 8$, which is less than $\int_{1}^{5} x dx = [\frac{x^2}{2}]_{1}^{5} = \frac{25-1}{2} = 12$.

Conversely, if $f(x) < x$, then $f^{-1}(x) > x$. This implies $\int_{1}^{5} f^{-1}(x) dx > \int_{1}^{5} x dx = 12$. Our result $16 > 12$ is consistent with this property.