Find a function f and a number a such that
x
2+∫ (f(t)/t^(5)) dt=6x^(−2)
a
f(x)=
a=
dF/dx = f(x) then
dF/dx = f(x)/x^5 = 6x^-2 - 2
see what you can do with that.
actually, I kind of mangled that. If
F(x) = ∫[a,x] f(t)/t^5 dt
then
F'(x) = f(x)/x^5 = -12/x^3
so, f(x) = -12x^3
F(x) = 6/x^2
So now we have
∫[a,x] f(t)/t^5 dt + 2 = F(x) - F(a) = 6/x^2 - 6/a^2 + 2 = 6/x^2
so
6/a^2 = 2
a = ±√3
To find a function f and a number a such that ∫(f(t)/t^(5)) dt = 6x^(-2), we need to find the antiderivative of f(t)/t^5 and then solve for f(x) and a.
To integrate f(t)/t^5, we can use the power rule of integration: ∫(x^n) dx = (x^(n+1))/(n+1) + C, where C is the constant of integration.
In this case, we have ∫(f(t)/t^5) dt. Let's apply the power rule with n = -5:
∫(f(t)/t^5) dt = (f(t) * t^(-5+1))/(-5+1) + C
= (f(t) * t^(-4))/-4 + C
= -f(t)/(4t^4) + C
Now, we have -f(t)/(4t^4) + C = 6x^(-2).
Since we want to find f(x) and a, we need to isolate f(x) from this equation. Let's substitute t with x and solve for f(x):
-f(x)/(4x^4) + C = 6x^(-2)
-f(x)/(4x^4) = 6x^(-2) - C
Multiplying both sides of the equation by 4x^4, we get:
-f(x) = 24 - 4Cx^4
Dividing by -1, we have:
f(x) = -24 + 4Cx^4
So, the function f(x) is -24 + 4Cx^4, where C is any constant.
Now, let's find the value of a:
We are given that ∫(f(t)/t^5) dt = 6x^(-2) from x = a to x.
Let's evaluate the integral using the found expression for f(x):
∫_(a)^(x) (f(t)/t^5) dt = 6x^(-2)
Substituting f(t) = -24 + 4Cx^4 and using the power rule for integration, we get:
∫_(a)^(x) (-24 + 4Ct^4)/t^5 dt = 6x^(-2)
Using the linearity property of integrals, we can split the integral:
-24∫_(a)^(x) (1/t^5) dt + 4C∫_(a)^(x) (t^4/t^5) dt = 6x^(-2)
Applying the power rule to the first integral, we have:
-24 * (-4)/4 * (1/t^4) | _(a)^(x) + 4C * (1/5) * (t^5/t^5) | _(a)^(x) = 6x^(-2)
Cancelling out some terms and simplifying, we get:
6/(x^2) = 6x^(-2)
Dividing both sides by 6, we have:
1/(x^2) = x^(-2)
Taking the reciprocal of both sides, we get:
(x^2) = 1/(x^2)
Multiplying both sides by (x^2), we have:
(x^4) = 1
Taking the square root of both sides, we get:
x^2 = ±1
Taking the square root again, we have:
x = ±1
Therefore, the value of a can be either 1 or -1.
So, the function f(x) = -24 + 4Cx^4, where C is any constant, and the number a can be either 1 or -1.
To find a function f(x) and a number a that satisfy the given equation, we will start by solving the integral on the left side:
∫ (f(t) / t^5) dt
To find the antiderivative of f(t) / t^5, we can use the power rule for integrals. By raising the power of t in the denominator by 1 and dividing by the new power, we get:
∫ (f(t) / t^5) dt = (f(t) / -4t^4) + C
where C is the constant of integration.
Now, substituting this result back into the original equation:
x^2 + ∫ (f(t) / t^5) dt = 6x^(-2)
x^2 + (f(t) / -4t^4) + C = 6x^(-2)
To simplify the equation, let's multiply both sides by t^4 to eliminate the fraction:
x^2 * t^4 - (f(t) / 4) + C * t^4 = 6
Now, let's differentiate both sides with respect to x to find the derivative of the left side:
d/dx [x^2 * t^4 - (f(t) / 4) + C * t^4] = d/dx [6]
2x * t^4 - (f'(t) / 4) + C * 4t^3 * dt/dx = 0
Simplifying this expression further, we get:
2x * t^4 - (f'(t) / 4) + 4Ct^3 * (dt/dx) = 0
Since we want the equation to hold for any value of x, the coefficient of each term should be equal to zero. Therefore:
2x = 0 (1) (Coefficient of t^4)
f'(t) / 4 = 0 (2) (Coefficient of f(t)/4)
4Ct^3 * (dt/dx) = 0 (3) (Coefficient of t^3 * (dt/dx))
From equation (1), we can conclude that x = 0.
From equation (2), f'(t) = 0, which means f(t) is a constant function.
Finally, from equation (3), since t is a variable and t^3 never equals zero, there is no condition on C.
Therefore, the function f(x) can be any constant function (f(t) = k, where k is a constant), and a = 0.
So, the desired function f(x) is f(x) = k, and the number a is a = 0.