1. ## definite integral

need help plz

2. Originally Posted by gracy
need help plz
In shorthand:
$\int_0^3 = \int_0^1 + \int_1^3$

so
$\int_1^3 = \int_0^3 - \int_0^1 = 4 - 7 = -3$

-Dan

3. Hello, gracy!

Make some sketches and use baby-talk . . .

$\int^1_0f(x)\,dx\:=\:7$ and $\int^3_0f(x)\,dx\:=\:4$
Find: $\int^3_1f(x)\,dx$

The first integral says: the area under $f(x)$ from 0 to 1 is $7$ square units.
The graph might look like this.
Code:
|
*
|   *
|     *
|  7   *
|
--+-------*----
|       1
|

The second integral says: the area under $f(x)$ from 0 to 3 is only $4.$

How did we lose some area?
Some of the area from 1 to 3 must be negative.
. . The graph must dip below the x-axis.

The graph might look like this.
Code:
|
*
|   *
|     *
|  7   *
|               3
--+-------*-------+--
|       1 *  -3 :
|           *   :
|               *

And that is why: . $\int^3_1f(x)\,dx\:=\:-3$