First, multiple choice questions
1. The following integral is correct ()
Shang ()
A. there is a maximum value of 0, and there is no minimum value.
B the maximum value is 0 and the minimum value is -323.
C. there is a minimum value of -323, and there is no maximum value.
D. there is neither a maximum nor a minimum.
[answer] b
[resolution] f (x) = 0x (t2-4t) dt =13t3-2t2x0 =13x3-2x2 (-1≤ x ≤ 5).
F' (x) = x2-4x, from f' (x) = 0, x = 0 or x = 4, as shown in the following table:
x(- 1，0)0(0，4)4(4，5)
f′(x)+0-0+
F(x)？ Maximum and minimum?
It can be seen that the maximum f (0) = 0 and the minimum f (4) =-323.
F (- 1) =-73，F (5) =-253。
∴ The maximum value is 0 and the minimum value is -323.
Second, fill in the blanks
1 1. Calculate the definite integral:
① 1- 1x2dx=________
②233x-2x2dx=________
③02|x2- 1|dx=________
④0-π2|sinx|dx=________
[Answer] 23; 436; 2; 1
[resolution] ①1-1x2dx =13x31-1= 23.
②233x-2x2dx=32x2+2x32=436。
③02 | x2- 1 | dx = 0 1( 1-x2)dx+ 12(x2- 1)dx
= x- 13x 3 10+ 13x 3-x 2 1 = 2。
[Answer] 1+π 2
13.(20 10? Shaanxi Institute, 13) If any point M(x, y) is taken from the rectangular area as shown in the figure, the probability that the point m takes the shaded part is _ _ _ _ _ _.
[Answer] 13
[Analysis] If the area of a rectangle is S 1 = 3, S Yin = 013x2xdx = x 310 =1,then P = S 1s Yin = 13.
14. It is known that f (x) = 3x2+2x+ 1. If 1- 1f (x) dx = 2f (a) holds, then a = _ _ _ _ _ _
[Answer]-1 or 13
[resolution] from the known f (x) = x3+x2+x, f (1) = 3, f (- 1) =- 1,
∴ 1- 1f(x)dx=f( 1)-f(- 1)=4，
∴2f(a)=4,∴f(a)=2.
That is 3A2+2A+ 1 = 2. The solution is A =- 1 or 13.
Third, answer questions.
15. Calculate the following definite integral:
( 1)052 xdx； (2)0 1(x2-2x)dx；
(3)02(4-2x)(4-x2)dx； (4) 12x2+2x-3xdx。
[resolution] (1) 052xdx = x250 = 25-0 = 25.
(2)0 1(x2-2x)dx = 0 1x2dx-0 12 xdx
= 13x 3 10-x 2 10 = 13- 1 =-23。
(3)02(4-2x)(4-x2)dx = 02( 16-8x-4x 2+2x 3)dx
= 16x-4x 2-43x 3+ 12x 420
=32- 16-323+8=403.
(4) 12 x2+2x-3xdx = 12x+2-3xdx
= 12 x2+2x-3 lnx 2 1 = 72-3 LN2。
16. Calculate the following definite integral:
[Analysis] (1) If f (x) = 12 sin2x, then f'(x)= cos2x.
= 12 1-32= 14(2-3).
(2) Let f (x) = x22+lnx+2x, then
f′(x)= x+ 1x+2。
∴23x+ 1x2dx=23x+ 1x+2dx
=F(3)-F(2)
=92+ln3+6- 12×4+ln2+4
=92+ln32。
(3) If f (x) = 32x2-cosx, then f' (x) = 3x+sinx.
17. Calculate the following definite integral:
( 1)0-4 | x+2 | dx；
(2) Given f (x) =, find the value of 3- 1f (x) dx.
[Resolution ]( 1)∫f(x)= | x+2 | =
∴0-4|x+2|dx=-4-2(x+2)dx+0-2(x+2)dx
=- 12 x2+2x-2-4+ 12 x2+2x 0-2
=2+2=4.
(2)∫f(x)= 1
∴3- 1f(x)dx=0- 1f(x)dx+0 1f(x)dx+ 12f(x)dx+23f(x)dx=0 1( 1-x)dx+ 12(x- 1)dx
=x-x22 10+x22-x2 1
= 12+ 12= 1.
18.( 1) Given f(a) = 0 1 (2ax2-a2x) dx, find the maximum value of f(a);
(2) it is known that f (x) = ax2+bx+c (a ≠ 0), and f (- 1) = 2, f' (0) = 0,01f (x) dx =-2, and find a, b, dx.
[Analysis] (1) Let f (x) = 23ax3- 12a2x2.
Then f' (x) = 2ax2-a2x.
∴f(a)=0 1(2ax2-a2x)dx
= F( 1)-F(0)= 23a- 12 a2
=- 12a-232+29
When a = 23, the maximum value of f(a) is 29.
(2)∵f(- 1)=2,∴a-b+c=2①
And ∵f'(x)= 2ax+b, ∴f'(0)= b = 0②.
And 01f (x) dx = 01(ax2+bx+c) dx.
Let f (x) =13ax3+12bx2+CX.
Then f' (x) = AX2+BX+C.
∴0 1f(x)dx=f( 1)-f(0)= 13a+ 12b+c=-2③
For solution ① ② ③, A = 6, b = 0 and c =-4.