Question: I am trying to compute the cumulative distribution function of a random variable $u$ that has the following density:

$$f(u) = \int\_{1}^\infty \frac{e^{-4uv}}{v^5}dv$$

for $u \gt 0$.

### What I've tried

I've tried computing this integral, giving a function $f(u)$, and then calculating $\int\_{-\infty}^x f(u) du$, obtaining the CDF of the density. But I always get some weird results from this integral. I know that there must be a better way to solve this using the fact that this is a joint density, but I don't know how to do it. Can someone please give me any hint or advice?

Answer: 
Your comment **"Now, I am not good with cars, so I do not know exactly what parts of the data here can help achieve better output"** is very insightful! You've correctly identified your problem, and the solution is not in the data or the model. The solution is finding out more about how cars accelerate.

I would advise you to find someone who does know a lot about cars, and ask them a lot of questions. You might learn things like:

* "Turbo charged engines always accelerate faster/slower than Kompressor charged engines"
* "If you want a car that accelerates fast, you would look for XXX and YYY but avoid ZZZ"
(You would then know to include those things in your model)
* "XXX increases acceleration up to a point but then levels off"
(Maybe you would then include XXX in your model as a polynomial term instead of a linear term)
* "For a front wheel drive car XXX is important but for a rear-wheel drive car XXX makes no difference"
(Then you might include an interaction term in your model)
* "Error of 0.8 is really close, that's very good accuracy" or "Error of 0.8 is really big, that's terrible accuracy"
(Then you would know how useful your model actually is)

I also do not know anything about cars, so I can't answer these questions. They may not even be the right questions! But for a real-world modelling task, understanding the real-world meaning of the data is never time wasted.
It is not clear what your question is, especially as the prediction looks amazingly accurate to me: 5.67 versus 5.8 is just an error of about 2%. Maybe you are misinterpreting the reported value of $R^2$ as a measure for accuracy, which it is not.

A nice thing about linear models is that they also estimate the accuracy of the prediction by means of a "prediction interval"\*). It estimates, under the assumption that the model is correct, an interval in which the response value will be observed 95% of the times for the given predictor values. In R, e.g., you can obtain it with

```
predict(model, data.frame(x1=val1, x2=val2, 
          ...), interval="predict")

```

This way, you can estimate the accuracy of the prediction even when the true value is unknown.

\*) Edit: in the original version of the answer, I called the "prediction interval" a "confidence interval", but as pointed out by @whuber, this term is reserved for parameter estimates.