Evaluate the definite integral.
9x e^(x^2)dx between (0,3)
u = x^2 Between (0,9)
du = 2x
du/2 = x
[9/2 e^u du ] from 0 t 9
[9/2 e^u ] from 0 t 9
9/2[e^9 -e^0]
9/2[e^9-1]= 36459.3776741
To evaluate the definite integral ∫[0,3] 9x e^(x^2) dx, we can use integration by substitution.
Let's start by making a substitution. We can let u = x^2. Then, du = 2x dx.
Next, we need to express everything in terms of u. We can rewrite the integral as ∫[0,3] 9x e^(u) (1/2x) du.
Simplifying further, we have ∫[0,3] 9/2 e^u du.
Now, we can integrate the expression with respect to u. The integral of e^u is simply e^u. So, the integral becomes (9/2) ∫[0,3] e^u du.
Evaluating the definite integral, we have (9/2) [e^u] evaluated from 0 to 3.
Plugging in the upper and lower limits, we get (9/2) [e^3 - e^0].
Since e^0 is equal to 1, the expression simplifies to (9/2) (e^3 - 1).
Therefore, the value of the definite integral ∫[0,3] 9x e^(x^2) dx is (9/2) (e^3 - 1).