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
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
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
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
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.
To find the value of A + B + E, we need to compare the given integral with the integration result -1/64 e^-4x [Ax^2 + Bx + E] + C and identify the corresponding coefficients.
Let's start by differentiating the expression -1/64 e^-4x [Ax^2 + Bx + E] + C to find the antiderivative:
d/dx (-1/64 e^-4x [Ax^2 + Bx + E] + C)
= -1/64 d/dx (e^-4x [Ax^2 + Bx + E]) + 0
= -1/64 [(d/dx e^-4x) (Ax^2 + Bx + E) + e^-4x (d/dx (Ax^2 + Bx + E))]
= -1/64 [-4e^-4x (Ax^2 + Bx + E) + e^-4x (2Ax + B)]
= -1/16 (e^-4x (-4Ax^2 - 4Bx - 4E + 2Ax + B))
= 1/4 (e^-4x (2Ax^2 - 2Ax + Bx - 4B - 4E))
Now, equate this with the given expression x^2 e^-4x:
x^2 e^-4x = 1/4 (e^-4x (2Ax^2 - 2Ax + Bx - 4B - 4E))
Comparing the coefficients on both sides of the equation, we get:
2Ax^2 - 2Ax + Bx - 4B - 4E = x^2
In order for the equation to hold true for all values of x, the coefficients of x^2, x, and the constant terms on both sides must be equal.
From the equation, we can conclude:
2A = 1 (coefficient of x^2)
-2A + B = 0 (coefficient of x)
-4B - 4E = 0 (constant term)
From the first equation, we solve for A:
2A = 1
A = 1/2
Substituting the value of A into the second equation, we can solve for B:
-2A + B = 0
-2(1/2) + B = 0
-1 + B = 0
B = 1
Finally, substituting the values of A and B into the third equation, we can solve for E:
-4B - 4E = 0
-4(1) - 4E = 0
-4 - 4E = 0
-4E = 4
E = -1
Therefore, A + B + E = 1/2 + 1 - 1 = 1/2.
However, none of the provided answer choices match this result, so there may be an error in the question or the answer choices.