Question

Integral
jika batas atas 3 batas bawah p (3x²-2x+2) = 112, maka nilai ½p adalah ....
a. 2
b. 1
c. -1
d. -2
e. -4

Answer 1

Berarti :
[3/3 x³ -2/2x² + 2x], batas atas 3, batas bawah p = 112
(3³ - (3)² + 2(3) - (p³ - p² + 2p)) = 112
(27 -9 + 6 ) - (p³ - p² + 2p) = 112
-p³ + p² - 2p = 112 - 24
p³ - p² + 2p = 88
p = -4

1/2p = 1/2(-4) = -2

Answer 2

Integral

p∫³=(3x²-2x+2)
=(x³-x²+2x)

batas atas = 3
batas bawah = p

p∫³ =((3)³-(3)²+2(3))-((p)³-(p)²+2(p)
=((27)-(9)+6))-((p)³-(p)²+2(p))
=(27-3)-(p³-p²+2p)
112 =(24)-(p³-p²+2p)

(p)³-(p)²+2(p)+112=24
p³-p²+2p=24-112
p³-p²+2p=-88
p³-p²+2p+88=0

sustitusikan
P=4
(-4)³-(-4)²+2(-4)=-88
(-64)-(16)-(8)=-88
-88=-88 (terbukti)

cara lain=

metode Horner

1 -1 +2 +88
-4 20 -88
(x=-4) ---------------- (+)
1 -5 22 0

(p+4)(p²-5p+22)=0
p=-4

1/2 p=1/2 (-4)=-2
Pilihan D.

Demikian Semoga Membantu dan Bermanfaat!