onsider the integral \({I}_{n}-\int_{0}^{1}x^n *{e}^{-2x}dx\)
1.Express \({I}_{n}\) in terms of\({I}_{n-1} for ....n>=1 \)
2. Hence, evaluate \(\int_{1}^{e} (lny/y)^3 dy \)
2-
Compute the definite integral:
integral_1^e (log^3(y))/y^3 dy
For the integrand (log^3(y))/y^3, integrate by parts, integral f dg = f g- integral g df, where
f = log^3(y), dg = 1/y^3 dy,
df = (3 log^2(y))/y dy, g = -1/(2 y^2):
= (-(log^3(y))/(2 y^2))|_1^e+1/2 integral_1^e (3 log^2(y))/y^3 dy
Evaluate the antiderivative at the limits and subtract.
(-(log^3(y))/(2 y^2))|_1^e = (-(log^3(e))/(2 e^2))-(-(log^3(1))/(2 1^2)) = -1/(2 e^2):
= -1/(2 e^2)+1/2 integral_1^e (3 log^2(y))/y^3 dy
Factor out constants:
= -1/(2 e^2)+3/2 integral_1^e (log^2(y))/y^3 dy
For the integrand (log^2(y))/y^3, integrate by parts, integral f dg = f g- integral g df, where
f = log^2(y), dg = 1/y^3 dy,
df = (2 log(y))/y dy, g = -1/(2 y^2):
= -1/(2 e^2)+(-(3 log^2(y))/(4 y^2))|_1^e+3/4 integral_1^e (2 log(y))/y^3 dy
Evaluate the antiderivative at the limits and subtract.
(-(3 log^2(y))/(4 y^2))|_1^e = (-(3 log^2(e))/(4 e^2))-(-(3 log^2(1))/(4 1^2)) = -3/(4 e^2):
= -5/(4 e^2)+3/4 integral_1^e (2 log(y))/y^3 dy
Factor out constants:
= -5/(4 e^2)+3/2 integral_1^e (log(y))/y^3 dy
For the integrand (log(y))/y^3, integrate by parts, integral f dg = f g- integral g df, where
f = log(y), dg = 1/y^3 dy,
df = 1/y dy, g = -1/(2 y^2):
= -5/(4 e^2)+(-(3 log(y))/(4 y^2))|_1^e+3/4 integral_1^e 1/y^3 dy
Evaluate the antiderivative at the limits and subtract.
(-(3 log(y))/(4 y^2))|_1^e = (-(3 log(e))/(4 e^2))-(-(3 log(1))/(4 1^2)) = -3/(4 e^2):
= -2/e^2+3/4 integral_1^e 1/y^3 dy
Apply the fundamental theorem of calculus.
The antiderivative of 1/y^3 is -1/(2 y^2):
= -2/e^2+(-3/(8 y^2))|_1^e
Evaluate the antiderivative at the limits and subtract.
(-3/(8 y^2))|_1^e = (-3/(8 e^2))-(-3/(8 1^2)) = 3/8-3/(8 e^2):
Answer: | = 3/8-19/(8 e^2)
