Question

In Exercises \(21-26,\) construct a function of the form \(y=\int^{x} f(t) d t+C\) that satisfies the given conditions. $$\frac{d y}{d x}=e^{x} \tan x,\( and \)y=0\( when \)x=8$$

Short answer

The solution to the problem is \(y = F(x) - F(8)\), where \(F(x)\) represents the integral function of \(e^{x} \tan x\).

Step-by-step solution

Integrate the function

Integrate the given function \(e^{x} \tan x\) with respect to \(x\), this will yield the original function \(y\). The integration can be computed as:\[\int e^{x} \tan x \, dx\]Unfortunately, the integral of \(e^{x} \tan x\) cannot be expressed in terms of elementary functions. Instead, we can represent it as an unspecified integral function, \(F(x)\). Therefore, the general form of \(y\) would be:\[y = F(x) + C\]

Solve for the constant C

To solve for \(C\), we will use the given condition that \(y = 0\) when \(x = 8\). Plug in these values into the expression \(y = F(x)+C\):\[0 = F(8) + C \]From this, it's clear that \(C = -F(8)\). Hence, the specific solution for the function \(y\) is \[y = F(x) - F(8)\]This indicates that \(y\) equals to the difference between the integral of \(e^{x} \tan x\) at points \(x\) and 8.