Need some help with this Integral Problem. Can someone show how they solved the problem below?

Question

oo
∫ e^-3x+2 . . dx =
0



possible solutions from the back of the book......

a. (1/2)e

b. 1/(2e)

c. 1/(3e)

d. (1/3)e^2

e. (1/2)e^3

or none of these ?

Answer 1

answer is d. integrate then substitute the absolute values

Answer 2

Int e^(-3x+2)dx = -1/3e^(-3x+2)+C
lim -1/3e^(-3x+2) as x=>+infinity is zero and at x=0 the value is-1/3e^2 so Int (0,+ infinity )=1/3e^2 (d)

Answer 3

? is spelled "pi" in English. i'm reading your statements as: arccos(x) + arcsin^2(y) = p*pi^2/4 arccos(x) * arcsin^2(y) = pi^2/sixteen resolve for arcsin^2(y) interior the 1st equation. i'm doing it this style so i'm able to get rid of the extra complicated expression first. arcsin^2(y) = p*pi^2/4 - arccos(x) Sub into the 2nd equation. arccos(x) * (p*pi^2/4 - arccos(x)) = pi^2/sixteen boost and simplify arccos(x) * p*pi^2/4 - arccos^2(x) = pi^2/sixteen To make existence a splash much less confusing, i'm going to make a short lived substituion. enable u = arccos(x) (p*pi^2/4)*u - u^2 = pi^2/sixteen Multiply each term via sixteen, just to do away with the fractions. 4pi^2p*u - 16u^2 = pi^2 pass all words to the left and turn the signs and indicators. -16u^2 + 4p*pi^2u - pi^2 = 0 16u^2 - 4p*pi^2u + pi^2 = 0 Use the quadratic answer to be certain the valid values of p. If Ax^2 + Bx + C = 0, then x = [-B +- sqrt(B^2 - 4AC)] / (2A) Sub in. u = [ 4p +- sqrt([4p*pi^2]^2 - 4[sixteen][pi^2]) ] / (2*sixteen) u = [ 4p +- sqrt(16p^2pi^4 - 64pi^2) ] / 32 The section interior the sq. root (sqrt) is call the discriminant. It tells us the style of recommendations. If the discriminant < 0, there are no longer any genuine recommendations. If the discriminant = 0, the comparable genuine answer looks two times. If the discriminant > 0, then there are 2 different genuine recommendations. So we'd like the discriminant >= 0. 16p^2pi^4 - 64pi^2 >= 0 Divide all words via sixteen pi^2. p^2pi^2 - 4 >= 0 resolve for p. p^2pi^2 >= 4 p^2 >= 4/pi^2 p <= -(2/pi) OR p >= 2/pi 2/pi =~ 0.63662

Answer 4

this is the easiet possible integral you can get.
No excuse to be lazy. I will not solve your question but here is an example.
is the 2 in your question a part of the exponent?
∫ e^-5x+1 . . dx =e^(-5x+1)/-5
we substitute limits, with b in place of oo
(e^-5b+1)/-5 - e^(0 +1)/-5
we take limit as b goes to infinity.
first term goes to 0.
answer e/5

Answer 5

integral is -1/3(e^-3x+2). then evaluate between zero and infinity, upper limit gives you nothing ,since e^-infininity is zero, lower limit gives u the negative of -1/3(e^-3x+2), substitute x=zero and you get d: (1/3)e^2

Answer 6

I also get (d). Use u-substitution, u = -3x+2. The whole solution falls out like clockwork.

Answer 7

the answer is d. (1/3)e^2