Question

How do you integrate `3t^2+2t+2` ?

Answer 1

`int (3t^2+2t+2) dt = t^3+t^2+2t+C`

Explanation:

I will use the following properties:

• `d/(dt) t^n = n t^(n-1)" "` if `n != 0`

• `int (d/(dt) f(t)) dt = f(t)+C" "` where `C` is a constant

• `int (f(t) + g(t)) dt = int f(t) dt + int g(t) dt`

• `int k f(t) dt = k int f(t) dt`

From the first two properties, we can deduce that if `n != -1` that:

`int t^n dt = 1/(n+1) t^(n+1)+C`

The third and fourth properties tell us that integration is a linear operator.

We can deduce that:

`int (3t^2+2t+2) dt = 3 int t^2 dt + 2 int t dt + 2 int 1 dt`

`color(white)(int (3t^2+2t+2) dt) = t^3+t^2+2t+C" "` where `C` is a constant.

Answer 2

`t^3 + t + 2t + C`

Explanation:

Given: Integrate `3t^2 + 2t + 2`

This is an indefinite integral. That means there are many solutions represented by the constant `C`.

`int (3t^2 + 2t + 2)dt = int 3t^2dt + int 2t dt + int 2 dt`

Since `int kdx = kx + C " and " int x^n dx = 1/(n+1) x^(n+1) + C, n != -1`:

`int 3t^2dt + int 2t dt + int 2 dt = 3/3t^3 + 2/2t + 2t + C`

`= t^3 + t + 2t + C`

Answer 3

See.

Explanation:

We have to integrate in respect to t.
`int(3t^2+2t+2)`
`=int3t^2+int2t+int2`
`=3intt^2+2intt+int2`[keeping the constant outside]
`=3×(t^3/3)+2×(t^2/2)+2t+c`[As it is a indefinite nominal having no limit]
`=t^3+t^2+2t+c`(ans)