Marc S. answered • 01/02/13
Former Boston University lecturer tutoring math and science
Hi Vishwa.
The area between a function and the x-axis (which is the same as the line y=0) is found from the integral of the function. What you need first are the values of x that bound the area you're interested in finding. One value is given to you: x=1. You can find the other value by solving the system of equations:
y=0 and y=xe(x^2) by using substitution to get: 0=xe(x^2).
From the zero product property, either x=0 or e(x^2) = 0. Attempting to solve the second equation results in no solutions. So now we have our lower bound of x=0 and our upper bound of x=1. The area is:
∫01 xe(x^2) dx.
Whoever came up with the answer e/2 probably ignored zero as the lower limit of integration. This is a safe thing to ignore when you're integrating polynomials. For example:
∫02 x2 dx is equal to F(2) - F(0)
where F(x) = x3/3, the antiderivative of x2.
So, ∫02 x2 dx =23/3 - 03/3 = 8/3 - 0 = 8/3.
See how the entire second antiderivative, F(0), can be ignored because you just subtract zero? Some students get so used to integrating polynomials that they always ignore a lower limit of zero, but with ∫01 xe(x^2) dx that's a big mistake, and it's the reason I can't confirm your solution.
To find ∫xe(x^2) dx, you need to use u-substitution. Instead of writing a complete solution here, I recommend an online tool called WolframAlpha. Try clicking http://www.wolframalpha.com/input/?i=area+bounded+by+x%3D1+and+y%3D0+and+y%3Dx*e^x^2 for the definite integral. For a step-by-step solution using u-substitution, click http://www.wolframalpha.com/input/?i=Integrate+y%3Dx*e^x^2 , and press the "Step-by-step solution" button. You'll need free WolframAlpha account, but this is an incredibly useful thing to have.
Marc S.
01/03/13
Matt L.
Be careful, Marc! It's misleading to speak of "the antiderative of x2," because there is no such thing --- antiderivatives aren't unique, so you can't use the definite article "the." Also, the fact that a student can ignore a lower limit of 0 in a definite integral has nothing to do with the integrand being a polynomial; it merely has to do with choosing an antiderivative F satisfying F(0)=0, i.e. with a y-intercept of 0. (In fact, by existence and uniqueness theorems from differential equations, this can always be done, no matter the integrand.) An example: when integrating e^x from 0 to 1, just choose the antiderivative to be e^x-1 rather than the "more standard" e^x. Then the lower limit of 0 can be "ignored," because this particular choice of antiderivative has a y-intercept of 0.
Your notation is also misleading: (e^x)^2 is one thing, equal to e^(2x), whereas e^(x^2) is another. (See Robert's post below.) The OP's post is ambiguous, of course, but you're wrong to say that x(e^x)^2 would be integrated by a u-substitution; it would require integration by parts.
01/03/13