In the graph it is the point with x- coordinate 1 and y-coordinate - 2
Answer: f ( 1 ) = - 2
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:
£5/£4
Step-by-step explanation:
£5 over £4 can be expressed as a fraction.
⇒ £5/£4
