Take an arbitrary interval [a, b], where a < b.
Compute the arc length L of y = f(x) over [a, b] :
[tex]L=\displaystyle\int_a^b\sqrt{1+\left(f'(x)\right)^2}\,\mathrm dx[/tex]
Now comptue the area A under the curve y = f(x) over [a, b] :
[tex]A=\displaystyle\int_a^bf(x)\,\mathrm dx[/tex]
We have
f (x) = 1/4 e ⁻ˣ + e ˣ → f ' (x) = -1/4 e ⁻ˣ + e ˣ
Then
√(1 + (f ' (x))²) = √(1 + (-1/4 e ⁻ˣ + e ˣ)²)
… = √(1 + 1/16 e ⁻²ˣ - 1/2 + e ²ˣ)
… = √(1/16 e ⁻²ˣ + 1/2 + e ²ˣ)
… = 1/4 √(e ⁻²ˣ + 8 + 16e ²ˣ)
… = 1/4 √((e ⁻ˣ + 4 e ˣ)²)
… = 1/4 (e ⁻ˣ + 4 e ˣ)
… = 1/4 e ⁻ˣ + e ˣ
… = f (x)
so both A = L for any choice of interval [a, b].
It is true that the arc length on any interval has the same value as the area under the curve.
The function is given as:
[tex]f(x) = \frac 14e^{-x} + e^x[/tex]
Differentiate the function
[tex]f'(x) = -\frac 14e^{-x} + e^x[/tex]
On any interval, the following must be true
[tex]f(x) =f'(x)[/tex]
and
[tex]f(x) = \sqrt{1 + (f'(x))^2}[/tex]
So, we have:
[tex]f(x) = \sqrt{1 + (-\frac 14e^{-x} + e^x)^2}[/tex]
Expand the exponents
[tex]f(x) = \sqrt{1 + (\frac{1}{16}e^{-2x} - \frac 12 + e^{2x})}[/tex]
Remove the bracket
[tex]f(x) = \sqrt{1 + \frac{1}{16}e^{-2x} - \frac 12 + e^{2x}}[/tex]
Evaluate the like terms
[tex]f(x) = \sqrt{\frac{1}{16}e^{-2x} + \frac 12 + e^{2x}}[/tex]
Multiply by 16/16
[tex]f(x) = \sqrt{\frac{16}{16}(\frac{1}{16}e^{-2x} + \frac 12 + e^{2x})}[/tex]
So, we have:
[tex]f(x) = \sqrt{\frac{1}{16}(e^{-2x} + 8 + 16e^{2x})}[/tex]
Take the square root of 1/16
[tex]f(x) = \frac{1}{4}\sqrt{e^{-2x} + 8 + 16e^{2x}}[/tex]
Express the radical as a perfect square
[tex]f(x) = \frac{1}{4}\sqrt{(e^{-x} + 4e^{x})^2}[/tex]
Evaluate the exponents
[tex]f(x) = \frac{1}{4} * (e^{-x} + 4e^{x})[/tex]
Evaluate the products
[tex]f(x) = \frac{1}{4}e^{-x} + e^{x}[/tex]
Hence, it has been proved that the arc length on any interval has the same value as the area under the curve.
Read more about areas at:
https://brainly.com/question/24487155
