Question

If each case, give an example of a continuous function \(\mathrm{f}\) satisfying the conditions stated for all real \(\mathrm{x}\), or else explain why there is no such function : (a) \(\int_{0}^{x} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{e}^{x}\) (b) \(\int_{0}^{x^{2}} f(t) d t=1-2^{x^{2}}\). (c) \(\int_{0}^{x} f(t) d t f^{2}(x)-1\).

Short answer

Question: Give an example of a continuous function that fulfills condition (a) and explain why it's not possible to find a function that fulfills condition (c). Answer: An example of a continuous function that fulfills condition (a) is \(f(t) = e^t\). This function fulfills the condition because its integral from 0 to x equals \(e^x\). On the other hand, it's not possible to find a continuous function that fulfills condition (c) because the left side of the equation (\(\int_{0}^{x} f(t) dt\)) is always non-negative, while the right side (\(f^{2}(x)-1\)) can have negative values for some x if the function is continuous. This contradiction indicates that no such continuous function exists for condition (c).

Step-by-step solution

(a) Find a function whose integral from 0 to x equals e^x

To find a continuous function f(t) satisfying the condition \(\int_{0}^{x} f(t) dt = e^{x}\), we will find the derivative of the function with respect to x. Using the Fundamental Theorem of Calculus, \(\frac{d}{dx} (\int_{0}^{x} f(t) dt) = f(x)\) So, \(\frac{d}{dx} (e^{x}) = e^{x}\), which means the function f(t) is \(f(t) = e^{t}\). This function is continuous for all real numbers.

(b) Find a function whose integral from 0 to x^2 equals 1-2^(x^2)

To find a continuous function f(t) satisfying the condition \(\int_{0}^{x^{2}} f(t) dt = 1-2^{x^{2}}\), we will again take the derivative of the given function with respect to x. \(\frac{d}{dx} (\int_{0}^{x^{2}} f(t) dt) = f(x^{2}) \cdot \frac{d}{dx} (x^{2})= 2xf(x^{2})\) And we know that, \(\frac{d}{dx} (1-2^{x^{2}}) = -2^{x^{2}} \cdot \ln(2) \cdot 2x\) So, \(f(x^{2}) = -\frac{1}{2} \ln(2) \cdot 2^{x^{2}}\), which implies that \(f(t) = -\frac{1}{2} \ln(2) \cdot 2^{t}\). This is a continuous function for all real numbers.

(c) Determine if a function exists that fulfills the given condition

The condition states \(\int_{0}^{x} f(t) dt = f^{2}(x)-1\). We can see that the left side of the equation is always non-negative since it's a definite integral, whereas the right side could be negative if \(f^2(x) < 1\). Additionally, if f(x) is continuous, it must be between 0 and 1 for some real x, which means there will be some negative values for the right side. This contradicts with the condition of the integral being non-negative, therefore there is no such continuous function that fulfills the condition (c).