1. ## Intergrating when f(x) is made up of two functions

Hi, I would be very grateful if anyone could help me with this problem. I can pretty much see how you're meant to do this, but one bit is causing a large problem, the result when you integrate the (x+1) function, in the answer given was rather vague as to why it behaved strangely at this point. Cheers!

2. You have to split the integral up as $\int_{t-1}^tf(x)\,dx = \int_{t-1}^1(x+1)\,dx + \int_1^t(2x^2-6x+6)\,dx$.

I'm not sure how the (x+1) part would cause trouble. You should get $\int_{t-1}^1(x+1)\,dx = \Bigl[\tfrac12(x+1)^2\Bigr]_{t-1}^1 = 2-\tfrac12t^2$.

3. Hello LHS
Originally Posted by LHS

Hi, I would be very grateful if anyone could help me with this problem. I can pretty much see how you're meant to do this, but one bit is causing a large problem, the result when you integrate the (x+1) function, in the answer given was rather vague as to why it behaved strangely at this point. Cheers!
$1\le t\le 2 \Rightarrow 0 \le (t-1)\le 1$

So the lower limit of the integral $\int_{t-1}^tf(x)\,dx$ lies between $0$ and $1$ inclusive, and the upper limit between $1$ and $2$ inclusive. So we shall need to split the integral into two ranges: $(t-1)$ to $1$ and $1$ to $t$, to take into account the change in the function $f(x)$ where $x = 1$. This gives:

$g(t) = \int_{t-1}^1(x+1)\,dx+\int_1^t(2x^2-6x+6)\,dx$

All you need to do now is to integrate each of these functions separately and evaluate them between their respective limits in the usual way, before simplifying the resulting polynomial in $t$.

You asked specifically about the $(x+1)$ function. This integrates to $\tfrac12x^2 + x$, of course, and when you put in the limits you'll get:

$\Big[\tfrac12x^2+x\Big]_{t-1}^1 = (\tfrac12+1) - \big(\tfrac12(t-1)^2+(t-1)\Big)$

When you combine this with the result of the second integral, I think the final answer is:

$g(t) = \tfrac23t^3-\tfrac72t^2+6t-\tfrac53$

Can you fill in the gaps?