suppose
x
S f(t) dt= x^2 - 2x + 1. Find f(x).
1
S = integral and 1 = lower level, x = upper
i don't understand what i'm supposed to find/what to do
did you mean "Find f(t)" ?
let f(t) = at^2 + bt + c
then the integral of f(t) by dt from t = 1 to t = x
= ax^2 + bx + c - (a + b + c)
= ax^1 + bx - a - b
comparing this with x^2 - 2x + 1 gives
a=1
b=-2 and
-a -b = 1 This last equation is true,
and c could be anything
so f(t) = t^2 - 2t + c
To find f(x), we need to evaluate the integral ∫f(t) dt = x^2 - 2x + 1, where the lower limit is 1 and the upper limit is x.
Let's break down the problem step by step:
1. Start with the given equation: ∫f(t) dt = x^2 - 2x + 1.
2. To find f(x), we need to find the antiderivative of the right side of the equation. The antiderivative of x^2 - 2x + 1 is obtained by integrating each term separately.
∫x^2 dt - ∫2x dt + ∫1 dt = x^3/3 - x^2 + x + C,
where C is the constant of integration.
3. Now, we have the equation: ∫f(t) dt = x^3/3 - x^2 + x + C.
4. To determine f(x), we need to equate the integrals on both sides of the equation.
∫f(t) dt = ∫f(t) dt = x^3/3 - x^2 + x + C.
5. Therefore, f(x) = x^3/3 - x^2 + x + C, where C is the constant of integration.
So, the function f(x) that satisfies the integral equation is f(x) = x^3/3 - x^2 + x + C.