Find the derivative of the following function using the appropriate form of the Fundamental Theorem of Calculus.
intergral s^2/(1+3s^4) ds from sqrtx to 1
F'(x)=?
Consider the function f(x) whose second derivative is f''(x)=8x+4sin(x). If f(0)=3 and f'(0)=2, what is f(x)?
To find the derivative of the given function, we will start by applying the Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus states that if a function f(x) is continuous on an interval [a, b], and F(x) is its antiderivative on that interval, then the definite integral of f(x) from a to b can be calculated as F(b) - F(a). In other words, the derivative of an integral can be obtained by evaluating the function at the upper limit of integration and subtracting the function evaluated at the lower limit of integration.
In this case, we have the integral:
∫(s^2 / (1 + 3s^4)) ds
We want to find the derivative of this integral with respect to x. To do so, we need to introduce the variables of integration. Let's rewrite the limits of integration in terms of x:
sqrt(x) to 1
Next, we need to insert s back in place of x in terms of integration. We can express s in terms of x as follows:
s = sqrt(x)
Now we have:
∫(s^2 / (1 + 3s^4)) ds from s = sqrt(x) to s = 1
To apply the Fundamental Theorem of Calculus, we need to find the antiderivative of the integrand. Let's do that:
Let u = 1 + 3s^4. Then, du = 12s^3 ds.
Rearranging the equation, we have s^3 ds = (1/12) du.
Now our integral becomes:
∫[(s^2 / (1 + 3s^4)) ds] = ∫[(s^2 / (u)) (1/12 du)]
Canceling out common factors, the integral simplifies to:
(1/12) ∫[(s^2 / u)] du
To integrate this, we rewrite s^2 in terms of u:
(1/12) ∫[(u - 1) / u] du
Splitting the integrand, we have:
(1/12) ∫[(u/u) - (1/u)] du
which simplifies to:
(1/12) ∫[(1 - (1/u))] du
Integrating each term, we get:
(1/12) [u - ln|u|] + C
Now, we substitute back u with 1 + 3s^4, and remember to evaluate the result at the upper and lower limits of integration:
F(x) = (1/12) [(1 + 3s^4) - ln|1 + 3s^4|] evaluated from s = sqrt(x) to s = 1
F(x) = (1/12) [(1 + 3(1)^4) - ln|1 + 3(1)^4|] - [(1/12) [(1 + 3(sqrt(x))^4) - ln|1 + 3(sqrt(x))^4|]
Finally, to find the derivative F'(x), we differentiate F(x) with respect to x:
F'(x) = (1/12) [0 - (3(1))^3 / (1 + 3(1)^4)] - [(1/12) [0 - (3(sqrt(x)))^3 / (1 + 3(sqrt(x))^4)]
Simplifying the above expression, we get:
F'(x) = -(3/4) - ( (3/4) (sqrt(x))^3 / (1 + 3(sqrt(x))^4) )
Therefore, the derivative of the given function is F'(x) = -(3/4) - ( (3/4) (sqrt(x))^3 / (1 + 3(sqrt(x))^4) ).