I can't do this question :-
Integrate the function f(x) = (x^3.e^-x^2)dx
I have tried many techniques but none seem to be working. can you please also give links of some useful techniques for questions such as these?
We have an integrand of: $\displaystyle x^3e^{-x^2}$
Use u-substitution by setting $\displaystyle u = x^2$
$\displaystyle du = 2xdx$
Now, the integrand looks like: $\displaystyle \frac{u}{2}e^{-u}$
Use integration by parts or tabular integration for this to get a result of: $\displaystyle \frac{-e^{-u}}{2}(u+1)$... subbing $\displaystyle x^2 = u$ takes us to a final answer of: $\displaystyle \frac{-e^{-x^2}}{2}(x^2+1)$
Hello, Altair!
I recommend "by parts" . . .$\displaystyle \int x^3e^{-x^2}\,dx$
. . $\displaystyle \begin{array}{ccccccc}u & = & x^2 & \quad & dv & = & xe^{-x^2}dx \\ du & = & 2x\,dx & \quad & v & = & -\frac{1}{2}e^{-x^2} \end{array}$
We have: .$\displaystyle -\frac{1}{2}x^2e^{-x^2} + \int xe^{-x^2}dx$
Hence: .$\displaystyle -\frac{1}{2}x^2e^{-x^2} - \frac{1}{2}e^{-x^2} + C \;=\;-\frac{1}{2}e^{-x^2}\left(x^2+1\right) + C$
A roller coaster at an amusement park starts out on a level track 50.0 m long and then goes up a 25.0-m incline at an angle of 30 degree to the horizontal. It then goes down a 15.0-m ramp with an incline of 40 degree to the horizontal. When the roller coaster has reached the bottom of the ramp, what is its displacement from its starting point?
I would suggest a new thread for a new problem. Folks are more likely to see it that way.
The displacement is how far it traveled from one point to another regardless of the 'hills and valleys'.
That's why it's called diplacement. How much distance it displaces.
It traveled horizontally for 50 feet. Then, using the law of cosines, it traveled
$\displaystyle x=25cos(30)$ and then $\displaystyle y=15cos(40)$
$\displaystyle 50+25cos(30)+15cos(40)$