Question

Ini soalnya.....
yg bisa kerjain yaa pke cara...

Answer 1

1) f(x) = 10 int (x√x) dx = int (10x^(3/2)) dx = 10 . 2/5 . x^(5/2) + C
f(x) = 4x^2 √x + C

f(1) = 4(1)^2 .√1 + C
=> 6 = 4 + C => C = 2

f(x) = 4x^2.√x + 2
f(4) = 4(4)^2 √4 + 2 = 4(16).2 + 2 = 128 + 2 = 130

2) f(x) = int (ax + 2(a - 1)) dx
f(x) = int (ax + 2a - 2) dx
f(x) = a/2 x^2 + 2ax - 2x + C

f(-2) = (a/2) (-2)^2 + 2a(-2) - 2(-2) + C
=> -12 = 2a - 4a + 4 + C
=> -12 = -2a + C + 4
=> 2a - C = 16 ................ (1)

f(2) = (a/2)(2)^2 + 2a(2) - 2(2) + C
=> 4 = 2a + 4a - 4 + C
=> 8 = 6a + C ............. (2)

eliminasi (1) dan (2)
2a - C = 16
6a + C = 8
-------------- +
8a = 24
a = 3

6a + C = 8
=> 6(3) + C = 8
=> C = -10

f(x) = (a/2) x^2 + 2ax - 2x + C
f(x) = (3/2)x^2 + 2(3)x - 2x + (-10)
f(x) = (3/2)x^2 + 4x - 10
f(1) = (3/2)(1)^2 + 4(1) - 10 = 3/2 + 4 - 10 = 3/2 - 6 = (3 - 12)/2 = -9/2 = -4 1/2