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3 years ago

Evaluate ∫ xe2x dx.

2 answers:

6 0

The formula is integral of (udv) = uv - integral of (vdu)
We use integration by parts by letting u = x and dv = (e^2x)dx
Then du = dx, and v = (1/2)(e^2x)
integral of x(e^2x)dx = (1/2)(x)(e^2x) - integral of (1/2)(e^2x)dx
= (1/2)(x)(e^2x) - (1/4)(e^2x) + C
Therefore the correct answer is C.

Helen [10]3 years ago

6 0

The answer is (1/2)xe^(2x) - (1/4)e^(2x) + C
Solution:
Since our given integrand is the product of the functions x and e^(2x), we can use the formula for integration by parts by choosing
u = xdv/dx = e^(2x)

By differentiating, we get
du/dx= 1
By integrating dv/dx= e^(2x), we have
v =∫e^(2x) dx = (1/2)e^(2x)

Then we substitute these values to the integration by parts formula:
∫ u(dv/dx) dx = uv −∫ v(du/dx) dx ∫ x e^(2x) dx
= (x) (1/2)e^(2x) - ∫ ((1/2) e^(2x)) (1) dx
= (1/2)xe^(2x) - (1/2)∫[e^(2x)] dx
= (1/2)xe^(2x) - (1/2) (1/2)e^(2x) + C
where c is the constant of integration.

Therefore, our simplified answer is now
<span> ∫ x e^(2x) dx = (1/2)xe^(2x) - (1/4)e^(2x) + C</span>

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Write the equation of a circle given the center (-4, 4) and raduis r = 5.

Len [333]

Answer:

A) $(x+4)^2+(y-4)^2=25$

Step-by-step explanation:

The equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where $(h,k)$ is the center and $r$ is the radius. If $(h,k)\rightarrow(-4,4)$ and $r=5$ , then:

$(x-h)^2+(y-k)^2=r^2\\\\(x-(-4))^2+(y-4)^2=5^2\\\\(x+4)^2+(y-4)^2=25$

Therefore, the equation of the circle is $(x+4)^2+(y-4)^2=25$

6 0

2 years ago

A 12 ft long playground is marked off into 1/5 ft long section for a game,how many sections are there

Mrac [35]

So its 12 divided by 1/5 so
12 x 1= 12
1 x 5= 5

That is 12 over 5 and u will divide the numerator by the denominator to get 2.4
Dont know if thts exactly right bcuz its been a while since i learned this but i hope it helps

4 0

3 years ago

The axis of symmetry for the function f(x) = –2x2 + 4x + 1 is the line x = 1. Where is the vertex of the function located?

egoroff_w [7]

Answer:

(1, 3)

Step-by-step explanation:

You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola. The first few steps are as follows. Set the parabola equal to 0 so you can solve for the vertex. Separate the x terms from the constant by moving the constant to the other side of the equals sign. The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out). Let's start there. The first 2 steps result in this polynomial:

$-2x^2+4x=-1$ . Now we factor out the -2:

$-2(x^2-2x)=-1$ . Now we complete the square. This process is to take half the linear term, square it, and add it to both sides. Our linear term is 2x. Half of 2 is 1, and 1 squared is 1. We add 1 into the set of parenthesis. But we actually added into the parenthesis is +1(-2). The -2 out front is a multiplier and we cannot ignore it. Adding in to both sides looks like this:

$-2(x^2-2x+1)=-1-2$ . Simplifying gives us this:

$-2(x^2-2x+1)=-3$

On the left we have created a perfect square binomial which reflects the h coordinate of the vertex. Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:

$-2(x-1)^2+3=y$

From this form,

$y=-a(x-h)^2+k$

you can determine the coordinates of the vertex to be (1, 3)

5 0

3 years ago

Read 2 more answers

There are 8 squares and 25% are red how many are blue

REY [17]

25% of 8 is 2, 8-2=6
2 are red, 6 are blue

5 0

3 years ago

Read 2 more answers

given examples of relations that have the following properties 1) relexive in some set A and symmetric but not transitive 2) equ

rodikova [14]

Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.

It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.

2) Let S=Z and define R = {(x,y) |x and y have the same parity}

i.e., x and y are either both even or both odd.

The parity relation is an equivalence relation.

a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.

b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.

c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.

Step-by-step explanation:

1) By definition,

a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.

That is, x works at the same place of x.

b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx

That is if x works at the same place y, then y works at the same place for x.

c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz

That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.

2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.

QED!

6 0

3 years ago