Question

Let \(f(x)=\int_{1}^{x} e^{t^{2}} d t\). Find the average value of \(f\) on the interval [0,1].

Short answer

The average value of the function \( f \) on the interval [0,1] is \( \int_{0}^{1}\int_{1}^{x} e^{t^{2}} dt dx \). But due to the complexity of the integral, the final result isn't an elementary function.

Step-by-step solution

Calculation of \( f(x) \)

The function \( f(x) \) is defined as \( f(x)=\int_{1}^{x} e^{t^{2}} dt \). But the antiderivative of \( e^{t^{2}} \) doesn't exist in elementary form. That means we can't provide a convenient algebraic expression for \( f(x) \). Therefore, \( f(x) \) remains as is (i.e., \( f(x)=\int_{1}^{x} e^{t^{2}} dt \)).

Calculate \( f(0) \) and \( f(1) \)

The goal is to calculate the average value on the interval [0,1], so we will need to find the values of \( f(0) \) and \( f(1) \). Evaluating the integrals, for \( f(0) \) use \( x=0 \) and for \( f(1) \) use \( x=1 \). Therefore, \( f(0) = \int_{1}^{0} e^{t^{2}} dt \) and \( f(1) = \int_{1}^{1} e^{t^{2}} dt = 0 \) as the limits are equal.

Calculate the average value of \( f \) on [0,1]

Since we don't have a convenient algebraic expression for \( f(x) \), the average value of \( f \) in the interval [0,1] will be evaluated using the definition of the average value of a function: \( \frac{1}{b-a} \int_{a}^{b} f(x) dx \), which gives \( \frac{1}{1-0} \int_{0}^{1}\int_{1}^{x} e^{t^{2}} dt dx \). So, the average value of \( f \) on [0,1] is \( \int_{0}^{1}\int_{1}^{x} e^{t^{2}} dt dx \).