Your answer is b. hope this helps :)
m1 = 63 m2 =117 m3=117 m5=117 m6=63 m7=117 m8=63
1/3(12x-24)=16
*3 *3
12x-24=48
+24 +24
12x=72
/12 /12
x=6
Answer:
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
Step-by-step explanation:
for
I= ∫x^n . e^ax dx
then using integration by parts we can define u and dv such that
I= ∫(x^n) . (e^ax dx) = ∫u . dv
where
u= x^n → du = n*x^(n-1) dx
dv= e^ax dx→ v = ∫e^ax dx = (e^ax) /a ( for a≠0 .when a=0 , v=∫1 dx= x)
then we know that
I= ∫u . dv = u*v - ∫v . du + C
( since d(u*v) = u*dv + v*du → u*dv = d(u*v) - v*du → ∫u*dv = ∫(d(u*v) - v*du) =
(u*v) - ∫v*du + C )
therefore
I= ∫u . dv = u*v - ∫v . du + C = (x^n)*(e^ax) /a - ∫ (e^ax) /a * n*x^(n-1) dx +C = = (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
Answer:
From -2<x<-1, the function F(X) is increasing. (B)
Really, it increases all from around -2.5<x<0.5
C is also the second answer, as it increases til around 2.5
It is decreasing from -4<x<-3. But increases right after. It then starts to slow down around x = 1 and go down again.
Which means (B) is your answer.
If you want to get fancy, its a polynomial and if you take the derivative for instantaneous rate, you will see f prime is increasing if you make an example function.
