Question

Find the interval in which \(F(x)=\int_{-1}^{x}\left(e^{t}-1\right)(2-t) d t, \quad(x>-1)\) is increasing.

Short answer

Answer: F(x) is increasing in the interval (0, 2).

Step-by-step solution

Differentiating F(x)

To find the derivative of F(x) with respect to x, we will use the Fundamental Theorem of Calculus, which states that if F(x) = \(\int_{a}^{x} f(t) dt\), then F'(x) = f(x). So, in our case, F'(x) = \((e^{x}-1)(2-x)\).

Find where F'(x) > 0

We want to find the interval in which F'(x) is positive, i.e., \((e^{x}-1)(2-x) > 0\). Let's analyze it in more detail. We have two factors \(e^x - 1\) and \(2-x\). Consider the factor \(e^x - 1\). Since \(e^x\) is an exponential function with its base greater than 1, it is always increasing for any value x. Also, it starts at 0 for x = 0. Therefore, \(e^x - 1\) will be greater than 0 for any x > 0 and lesser than 0 for any x < 0. Now let's move to the factor \(2 - x\). This is a decreasing linear function because the coefficient of x is negative. Thus, \(2 - x\) is greater than 0 for any x < 2 and lesser than 0 for any x > 2.

Determine the intervals where the product is positive

Combining our analysis of the two factors, we now need to find when their product, \((e^{x}-1)(2-x)\), is greater than 0. We can list the different cases for the signs of each factor. 1. When both factors are positive: \(e^x - 1 > 0\) and \(2 - x > 0\). This means x > 0 and x < 2, so the interval where both factors are positive is \((0, 2)\). 2. When both factors are negative: \(e^x - 1 < 0\) and \(2 - x < 0\). This means x < 0 and x > 2, which means there is no common interval where both factors are negative. As we can see, the product is positive only in the first case, so the interval in which F(x) is increasing is \((0, 2)\).