Question

Estimate the value of the number \({\rm{c}}\) such that the area under the curve \({\rm{y = sinhcx}}\) between \({\rm{x = 0}}\) and \(x = 1\) is equal to 1.

Short answer

The value of the number\(c\)is \(1.62\).

Step-by-step solution

Step 1: Remembering the formula

Considering that,

Area under curve\({\rm{y = f(x)}}\)between \(x{\rm{ }} = {\rm{ }}a\)and \({\rm{x = b}}\) is

\(\int_a^b | f(x)|dx\)

Finding Area under the curve

Given, Area under curve\(y = \sinh cx\)between \(x = 0\) and \(x = 1\) is

\(\int_0^1 | \sinh (cx)|dx\)

So, when we have \(c < 0\), \(\sinh (cx)\) is below the \(x - axis\)in the interval \((0,1)\).

Hence, area under the curve will be \(\infty \).

Evaluating further

When \(c < 0\)and \(|\sinh (cx)| = \sinh (cx)\)

Then, \(\int_{\rm{0}}^{\rm{1}} {{\rm{sinh}}} {\rm{(cx)dx}}\)

Use the formula : \(\int {{\rm{sinh}}} {\rm{(x)dx = cosh(x) + C}}\)

\(\begin{aligned}{c} &= \left( {\frac{{\cosh (cx)}}{c}} \right)_0^1\\ &= \frac{{\cosh (c)}}{c} - \frac{{\cosh (0)}}{c}\\ &= \frac{{\cosh (c) - 1}}{c}\end{aligned}\)

Point of intersection

Area under the curve will be \({\rm{1}}\), when

\(\frac{{{\rm{cosh(c) - 1}}}}{{\rm{c}}}{\rm{ = 1}}\)

Plot the graph for \({\rm{y = }}\frac{{{\rm{cosh(x) - 1}}}}{{\rm{x}}}\)

And the line \({\rm{y = 1}}\)

The \({\rm{x}}\)-coordinate of the point of intersection is the result value of \(c\).

Plotting the required graph

We can see by the graph, there is only one point of intersection.

We will zoom the x-coordinate of the point of intersection.

Final result

We can see that by the graph, \({\rm{x}}\)-coordinate of the point of the intersection is \({\rm{1}}{\rm{.62}}\)

Hence, \({\rm{c}} \approx {\rm{1}}{\rm{.62}}\)