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)
The answer is 7/4, 1 3/4 or 1.75
10 times larger than a thousand. Also one decimal place larger
Answer:
<em>Step-by-step explanation:</em>
<em>Step 1. (The capture) Capture a sample total counted =958, and them back tagged counted=38,total tagged=56
</em>
<em>The idea is to estimate the proportion p = m/N of tagged =102.
</em>
<em>Step 2. (The recapture) After everything has settled down, capture a new sample of n fish. Count the number of tagged fish. Suppose that k of them are tagged.
</em>
<em>It is reasonable that, k/n would be a good estimante for p = m/N.
</em>
<em>Accordingly, for an estimate N of N, se solve the equation m/N= k/n.
</em>
You can’t technically solve it because there’s no question