get out your integration by parts. Tedious, but not difficult.
Start with
u = x^2, du = 2xdx
dv = e^(-4x) dx, v = -1/4 e^(-4x)
∫x^2 e^(-4x) dx = -1/4 x^2 e^(-4x) + 1/2 ∫x e^(-4x) dx
Now let
u = x, du = dx
dv = e^(-4x) dx, v = -1/4 e^(-4x)
∫x e^(-4x) dx = -1/4 x e^(-4x) + 1/4 ∫ e^(-4x) dx
Put it all together now, and collect terms.
You can verify your answer at wolframalpha.com
If the integral of x^2 e^-4x dx=-1/64 e^-4x [Ax^2+Bx+E]+C, then the value of A + B + E is
a) -14
b) 10
c) 12
d) 26
4 answers
I have a lot of homework and Calculus test and I have "A" as grade including in the exams but there are certain problems of Calculus that confuse me too much and when I see them I realize that they were not difficult and I can understand them. I can not have any results from the options that the teacher gives me in this example. I do not know what is happening to me. Help me with this one too. Thanks in advance and sorry for the inconvenience
Hmmmph. Too bad you can't at least show some work. Putting together what I showed you above,
∫x^2 e^(-4x) dx = -1/4 x^2 e^(-4x) + 1/2 ∫x e^(-4x) dx
= -1/4 x^2 e^(-4x) + 1/2 (-1/4 x e^(-4x) + 1/4 ∫ e^(-4x) dx)
= -1/4 x^2 e^(-4x) + 1/2 (-1/4 x e^(-4x) + 1/4 (-1/4 e^(-4x)))
= e^(-4x) (-1/4 x^2 - 1/8 x - 1/32)
= -1/32 e^(-4x) (8x^2 + 4x + 1)
= -1/64 e^(-4x) (16x^2 + 8x + 2)
16+8+2 = 26
Now where did you get lost in that? It's just Algebra I
∫x^2 e^(-4x) dx = -1/4 x^2 e^(-4x) + 1/2 ∫x e^(-4x) dx
= -1/4 x^2 e^(-4x) + 1/2 (-1/4 x e^(-4x) + 1/4 ∫ e^(-4x) dx)
= -1/4 x^2 e^(-4x) + 1/2 (-1/4 x e^(-4x) + 1/4 (-1/4 e^(-4x)))
= e^(-4x) (-1/4 x^2 - 1/8 x - 1/32)
= -1/32 e^(-4x) (8x^2 + 4x + 1)
= -1/64 e^(-4x) (16x^2 + 8x + 2)
16+8+2 = 26
Now where did you get lost in that? It's just Algebra I
or , assume
y = -1/64 (e^(-4x) ) [Ax^2+Bx+E]+C <----- your given integral
dy= -(1/64)[ e^(-4x)(2Ax + B) - 4e^(-4x)(Ax^2 + Bx + E)] + 0
= -(1/64)(e^(-4x))[ (2Ax + B - 4Ax^2 - 4Bx - 4E]
= -e^(4x) [Ax/32 + B/64 - Ax^2/16 - Bx/16 - E/16]
= e^(4x) [-Ax/32 - B/64 + Ax^2/16 + Bx/16 + E/16]
compare to
x^2 e^(-4x)
conclusion:
Ax^2/16 = x^2
A/16 = 1
A = 16
-Ax/32 + Bx/16 = 0 , there was no x term
B/16 = A/32 , B/16 = 1/2 -------> B = 8
- B/64 + E/16 = 0 , there was no constant term
E/16 = 8/64
E = 2
then the given -1/64 (e^(-4x) ) [Ax^2+Bx+E]
= -1/64 (e^(-4x) ) [16x^2 + 8x + 2]
A+B+E = 26
proof:
www.wolframalpha.com/input/?i=differentiate+-1%2F64+(e%5E(-4x)+)+%5B16x%5E2+%2B+8x+%2B+2%5D+wrt+x
Did it this way just for the sheer fun of it.
y = -1/64 (e^(-4x) ) [Ax^2+Bx+E]+C <----- your given integral
dy= -(1/64)[ e^(-4x)(2Ax + B) - 4e^(-4x)(Ax^2 + Bx + E)] + 0
= -(1/64)(e^(-4x))[ (2Ax + B - 4Ax^2 - 4Bx - 4E]
= -e^(4x) [Ax/32 + B/64 - Ax^2/16 - Bx/16 - E/16]
= e^(4x) [-Ax/32 - B/64 + Ax^2/16 + Bx/16 + E/16]
compare to
x^2 e^(-4x)
conclusion:
Ax^2/16 = x^2
A/16 = 1
A = 16
-Ax/32 + Bx/16 = 0 , there was no x term
B/16 = A/32 , B/16 = 1/2 -------> B = 8
- B/64 + E/16 = 0 , there was no constant term
E/16 = 8/64
E = 2
then the given -1/64 (e^(-4x) ) [Ax^2+Bx+E]
= -1/64 (e^(-4x) ) [16x^2 + 8x + 2]
A+B+E = 26
proof:
www.wolframalpha.com/input/?i=differentiate+-1%2F64+(e%5E(-4x)+)+%5B16x%5E2+%2B+8x+%2B+2%5D+wrt+x
Did it this way just for the sheer fun of it.
1 answer