Antiderivative Help???

Question

[4sqrt(t^3)-3t]/sqrt(t^3)
x(9)=7
find the specific antiderivative and what is the value of x (all i know is that the answer is a number)

Answer 1

f(x) = [4sqrt(t^3) - 3t] / sqrt(t^3)

First, make everything have an exponent. Note that taking the square root is the same as taking it to the power of (1/2).

f(x) = [4[t^3]^(1/2) - 3t] / [t^3]^(1/2)

We can multiply exponents which are a power to a power.

f(x) = [4t^(3/2) - 3t] / t^(3/2)

Now, split into two separate fractions.

f(x) = 4t^(3/2)/t^(3/2) - 3t/t^(3/2)

Note we can subtract exponents when dividing.

f(x) = 4 - 3t^1/t^(3/2)

f(x) = 4 - 3t^(1 - 3/2)

f(x) = 4 - 3t^(-1/2)

Apply the reverse power rule, and denote the general antiderivative to be F(x). Remember that the antiderivative of x^n (for n not equal to -1) is (1/[n + 1])x^(n + 1).

F(x) = 4t - 3[ 2t^(1/2) ] + C

F(x) = 4t - 6t^(1/2) + C

BUT, we know F(9) = 7, so

F(9) = 4(9) - 6[9]^(1/2) + C = 7
4(9) - 6[9]^(1/2) + C = 7
36 - 6(3) + C = 7
36 - 18 + C = 7
18 + C = 7
C = 7 - 18 = -11

Therefore, now that we have C,

F(x) = 4t - 6t^(1/2) - 11

Answer 2

To integrate this, let's start out by simplifying the expression to something a little more friendly. Usually a good way to go if you have something like sqrt(x) is to multiply by 1, using 1 = sqrt(x)/sqrt(x), so in this case, let's multiply by sqrt(t^3)/sqrt(t^3):

(4sqrt(t^3) - 3t)/(sqrt(t^3)) (sqrt(t^3))/(sqrt(t^3))
= (4t^3 - 3t(sqrt(t^3))/t^3

Remembering that multiplying like bases raised to powers yields the base raised to the sum of the powers, and dividing yields the same base raised to the difference of the powers, we can simply this more by writing:


(4t^3 - 3t^(5/2))/t^3 = 4 - 3t^(-.5)


That's better! Now we can integrate it easily using the power rule:

int(4 - 3t^(-.5)) = 4t - 6t^(.5) + C

Your given initial condition is f(9) = 7, so the specific antiderivative is

7 = 4(9) - 6(9^(.5)) + C, C = -11

so 4t - 6t^(.5) - 11

Answer 3

[4sqrt(t^3)-3t]/sqrt(t^3) x(9)=7
=4-3t/t^3/2 = 4 - 3t^-1/2
intergral = 4t - (3/2)t^1/2 +C

The x(9) = 7 is confusing since the antiderivative is a function of t. If the answer is a number, then we must jnow the limits of integration so that the C will drop out.

I'm going to assume the limits of integration are 7 to 9.
Then 4(9) -1.5(3) - (4(7) - 1.5sqrt(7)) = 3.5 +1.5sqrt(7).

Answer 4

Well, the antiderivative is the same of integration :

integrate[4sqrt(t^3) - 3t] / sqrt(t^3)

integrate[ 4 - 3t / t^3/2]dt

integrate[4dt - 3t^-1/2dt]

then, that's an easy integration :

4t - 6t^1/2 + c = F(t)

the : F(9) = 7

so : 4*9 - 6*3 + c = 7 >>>> c = -11

F(t) = 4t - 6t^1/2 - 11

That's what I understand from the problem, hope that's the answer.

Answer 5

Take the antiderivative of [4√t³ - 3t] / √t³.

First break it up into separate fractions.

[4√t³ - 3t] / √t³ = 4 - 3/√t

Now we can integrate.

∫{[4√t³ - 3t] / √t³}dt
= ∫(4 - 3/√t)dt = 4t - 6√t + C

But
x(9) = 4t - 6√t + C = 4*9 - 6√9 + C = 7
x(9) = 36 - 6*3 + C = 36 - 18 + C = 18 + C = 7
C = -11

So
x(t) = 4t - 6√t - 11

Answer 6

4 sqrt(t^3)/sqrt(t^3) = 4

3t / sqrt(t^3) = 3t/t^1.5 = 3t^-0.5

The integral of it with respect to dt id:

4t - 2*3*t^0.5 + c = 4t - 6 sqrt(t) + c

let's put t=9
7=4*9- 6 sqrt(9) + c = 36 - 18 + c = 18+c

c = -11

x = 4t - 6 sqrt(t) - 11

Answer 7

x
= ∫[4sqrt(t^3)-3t]/sqrt(t^3) dt
= 4t - 6√t + c

Plug in x(9) = 7 and solve for c,
c = -11

x = 4t - 6√t - 11