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:
y value
Step-by-step explanation:
Answer:
1.5 bars of granola per student
Step-by-step explanation:
Number of students = 4
Number of granola bars = 6
Four students share 6 granola bars equally.
Each student gets = 6 ÷ 4
= 1 1/2
= 1.5 bars of granola per student
Each student's share is 1.5 granola bars.
Answer:
0.75
Step-by-step explanation:
Outlier - A value that "lies outside" than most of the other values.
The outlier is therefore 12, because it is not in the general range of values, which is on average around 29.
