a) Is ∫[-1 to 1]e^x^3 dx positive, negative, or zero? Explain.
I think is positive but i don't know how to explain it.
b) Explain why 0 < ∫[0 to 1] e^x^2 dx <3.
Can you type this equation or whatever in a better format? What branch of math does this problem come from?
Rachel
It's definitely a calculus problem, but some of the symbols didn't translate to ASCII. You'll need to give a brief description of the problem for the symbols you can't type.
Is a) the integral of e^(x^3)dx? on the interval [-1,1]? If so, then you should see that the function is positive on that interval, thus the area is positive.
Part b) appears to put a bound on the definite integral e^(x^2) on the interval [0,1]. f(x)=e^(x^2) is a monotonic increasing funtion on R, has a min of 1 on that interval, and a max of e on that interval The area is <=e<3, and the area is positive since the function is positive there. Thus 0<f(x)<3 for x in [0,1]
After checking my work I see I stated the function for b) wrong. f(x) is monotonic increasing on the interval [0,1], not on R.
Both problems are asking you to make determinations about the area of some exponential function. In b) e^(x^2) is symmetric about the x-axis, but it's increasing on the interval [0,1]. To see this, simply differentiate the function and you'll see it's slope is increasing at every point there.
a) To determine if the integral ∫[-1 to 1]e^x^3 dx is positive, negative, or zero, we need to evaluate the integral. However, since computing this exact integral may be difficult, we can use some reasoning to estimate the result.
The integrand e^x^3 is always positive, because the exponential function is always positive. Therefore, the signed area under the curve will also be positive.
Additionally, the interval of integration [-1 to 1] does not change the sign of the function. So, regardless of the interval, the integral will still be positive.
Therefore, ∫[-1 to 1]e^x^3 dx is positive.
b) The integral ∫[0 to 1]e^x^2 dx can be evaluated by finding the antiderivative of e^x^2 and evaluating it at the limits of integration. However, to understand why 0 < ∫[0 to 1]e^x^2 dx < 3, we can analyze the function itself.
The function f(x) = e^x^2 is always positive for all real values of x. This means that the area under the curve will also be positive.
Furthermore, we can observe that the function is increasing on the interval [0 to 1]. This can be proven by taking the derivative of f(x) = e^x^2 and showing that it is positive on the interval.
Since the function is always positive and increasing on the interval [0 to 1], the integral ∫[0 to 1]e^x^2 dx will also be positive. However, it is bounded by the maximum value of the function, which is e, and the minimum value of the function, which is 1.
Therefore, 0 < ∫[0 to 1]e^x^2 dx < 3.