That should work.qspeechc said:
You just did the sub u=x^2 a couple of steps ago.qspeechc said:
qspeechc said:
), which I do.To solve this integration problem, we can use the substitution method. Let u = x^2, then du = 2x dx. We can rewrite the integral as $\frac{1}{2} \int_{0}^{\infty}u.e^{-u}\mathrm{d}u$. We can then use integration by parts to solve for the antiderivative, which is $\frac{1}{2}(-e^{-u}(u+1))$. Substituting back in for u, we get the final answer of $\frac{1}{2}(-e^{-x^2}(x^2+1))$.
The upper limit of infinity in this integration problem signifies that we are integrating over an unbounded interval. In other words, we are finding the area under the curve of the function from 0 to infinity.
Yes, there are other methods that can be used to solve this integration problem. Some examples include using partial fractions or trigonometric substitutions.
Yes, we can check the solution by taking the derivative of the antiderivative we found and seeing if it matches the original integrand. In this case, the derivative of $\frac{1}{2}(-e^{-x^2}(x^2+1))$ is indeed $x^3.e^{-x^2}$, so our solution is correct.
This integration problem can be solved by hand, as shown in the previous steps. However, a computer can also be used to solve it numerically, which may be necessary for more complex integration problems.