Question

How do you evaluate the definite integral `int (t^2-2)` from `[-1,1]`?

Answer 1

`-10/3`

Explanation:

You would simply apply The Fundamental Theorem of Calculus, Part II. This (simplified) states `int_a^bf(x)dx = F(b) - F(a)` where `F(x) = int f(x)dx`. Since `int (t^2-2) dt = t^3/3-2t` you can apply the theorem.

`int_-1^1(t^2-2)dt = 1^3/3-2(1)-((-1)^3/3-2(-1)) = 1/3-2+1/3-2 =2/3-4 = -10/3`

Answer 2

I have assumed that you are using the Fundamental Theorem of Calculus. If you are using the limit definition of definite integral,see another answer.

Explanation:

The integrand `f(t) = t^2-2` is an even function and we are integrating from `-a` to `a`.

Some instructors and some students appreciate the fact that for `f` an even function that is integrable on `[-a,a]`, we have

`int_-a^a f(x) dx = 2int_0^a f(x) dx`.

`int_-1^1 (t^2-2) dt = 2int_0^1(t^2-2) dt`

`= 2[t^3/3-2t]_0^1`

`= 2[1/3-2] = 2(-5/3) = -10/3`

Answer 3

If you are using a limit definition to do this, here is an answer.

Explanation:

Here is a limit definition of the definite integral. (I'd guess it's the one you are using.) I will use what I think is somewhat standard notation in US textbooks.

.`int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax`.

Where, for each positive integer `n`, we let `Deltax = (b-a)/n`

And for `i=1,2,3, . . . ,n`, we let `x_i = a+iDeltax`. (These `x_i` are the right endpoints of the subintervals.)

I prefer to do this type of problem one small step at a time.

In this question, the variable name is `t`. We'll use that.

`int_-1^1 (t^2-2) dt`.

Find `Delta t`

For each `n`, we get

`Deltat = (b-a)/n = (1-(-1))/n = 2/n`

Find `t_i`

And `t_i = a+iDeltat = -1+i2/n = -1+(2i)/n`

Find `f(t_i)`

`f(t_i) = (t_i)^2+2 = (-1+(2i)/n)^2 - 2`

`= (1-(4i)/n +(4i^2)/n^2) - 2`

`= -1 - (4i)/n+(4i^2)/n^2`

Find and simplify `sum_(i=1)^n f(t_i)Deltat` in order to evaluate the sums.

`sum_(i=1)^n f(t_i)Deltat = sum_(i=1)^n ( -1 - (4i)/n+(4i^2)/n^2) 2/n`

`= sum_(i=1)^n( (-2)/n - (8i)/n^2+(8i^2)/n^3)`

`=sum_(i=1)^n ((-2)/n) - sum_(i=1)^n ((8i)/n^2) + sum_(i=1)^n ((8i^2)/n^3)`

`=(-2)/n sum_(i=1)^n (1) - 8/n^2 sum_(i=1)^n(i)+8/n^3 sum_(i=1)^n(i^2)`

Evaluate the sums

`= (-2)/n (n) -8/n^2((n(n+1))/2) +8/n^3 ((n(n+1)(2n+1))/6)`

(We used summation formulas for the sums in the previous step.)

Rewrite before finding the limit

`sum_(i=1)^n f(t_i)Deltax = (-2)/n (n) -8/n^2((n(n+1))/2) +8/n^3 ((n(n+1)(2n+1))/6)`

`= -2 -4((n(n+1))/n^2) +4/3 ((n(n+1)(2n+1))/n^3)`

Now we need to evaluate the limit as `nrarroo`.

`lim_(nrarroo) ((n(n+1))/n^2) = lim_(nrarroo) (n/n*(n+1)/n) = 1`

`lim_(nrarroo) ((n(n+1)(2n+1))/n^3) = lim_(nrarroo)(n/n * (n+1)/n * (2n+1)/n) = 2`

To finish the calculation, we have

`int_0^1 (t^2 -2) dt = lim_(nrarroo) (-2 -4((n(n+1))/n^2) +4/3 ((n(n+1)(2n+1))/n^3))`

`= -2-4(1)+4/3(2)`

`= -6+8/3 = (-18+6)/3 = -10/3`.