Question

Nilai p yang memenuhi batas atas p batas bawah 2 (x³-16x) dx = -36 adalah ....
a. 4
b. 6
c. 8
d. 12
e. 16

Answer 1

Integral

f(x)=(x³-16x)dx=-36

₂∫ᵖ= 1/4x⁴-16/2x²
= 1/4x⁴-8x²

nilai p =
batas atas=p
batas bawah=2
=((1/4(p)⁴-8(p)²)-((1/4(2)⁴-8(2)²)=-36
=((1/4(p)⁴-8(p)²=((1/4(16)-8(4))+(-36)
=((4-32-36))
=((-28-36))
=((-64))

1/4(p)⁴-8(p)²=-64
1/4(p)⁴-8(p)²+64=0

kita sustitusikan p=4
1/4(4)⁴-8(4)²=-64
1/4(256)-8(16)=-64
64-128=-64 (terbukti)

Jawaban
Pilihan A.4

cara lain=

metode Horner

1/4 0 -8 0 +64
x=4 1 4 -16 -64
--------------------- (+)
1/4 1 -4 -16 0

(p-4)(1/4p³+p²-4p-16)=0
(1/4p⁴+p³-4p²-16p)-(p³+4p²-16p-64)=0
1/4p⁴+p³-p³-4p²-4p²-16p+16p+64=0
1/4p⁴-8p²+64=0 (terbukti)

p=4
-----

Demikian Semoga Membantu dan Bermanfaat!