Attached.

2. The three subintervals are [2;5] , [5;10] , and [10;12] . Using the trapezoidal rule, we get
f (2) + f (5)
f (5) + f (10)
f (12) + f (10)
(5 − 2) +
(10 − 5) +
(12 − 10) =
2
2
2

12

 f ( x)dx 
2

100 + 40
40 − 120
−120 − 150
=3
+5
+2
= 210 − 200 − 270 = −260
2
2
2

.

5. By the Newton-Leibniz formula we obtain
9

 f ( x)dx = f (9) − f (3) = 6 − 3 = 3 .
3

7. By the additive interval property we have
10



3

4

10

2

3

4

f ( x)dx =  f ( x)dx +  f ( x)dx +  f ( x )dx .

2

The same additive property says that
3

2

2

2

3

2

 f ( x)dx +  f ( x)dx =  f ( x)dx = 0 .
3

It follows that

2

 f ( x)dx = −  f ( x)dx = 6 and
2

3

10



10

3

4

2

2

3

f ( x)dx =  f ( x)dx −  f ( x)dx −  f ( x)dx = 18 − 6 − 1 = 11 .

4

8. Note that the function is differentiable on [−1;5] .
(a) The derivative is non-positive on (−1; 4) so the function does not increase on the
interval, f ( x)  0 x  (−1; 4) . From the other side the derivative is positive on (4;5) so the
function increases on the interval, f ( x)  0 x  (4;5) . Hence f ( x) has local maxima at the
points x = −1 and x = 5 , and local minima at the point x = 4 .
(b) The point x = 4 is the only local minimum on the segment [−1;5] , therefore f ( x)
achieves its absolute minimum value at the point.
5

(c) Let us present the integral

 f ( x)dx = f (5) − f (−1)

as

−1

5



−1

0

2



4

5

−1

0

2



4

f ( x)dx =  f ( x)dx +  f ( x)dx +  f ( x)dx +  f ( ...