jodecy said:
The notation ∫f(x)dx[b, a] represents the definite integral of the function f(x) over the interval [b, a]. This means that the integral is being evaluated from the lower limit b to the upper limit a.
To prove that ∫f(x)dx[b, a] =∫f(x−c)dx [b+c, a+c] means to show that the definite integral of the function f(x) over the interval [b, a] is equivalent to the definite integral of the function f(x-c) over the interval [b+c, a+c].
Proving that ∫f(x)dx[b, a] =∫f(x−c)dx [b+c, a+c] is important because it demonstrates a property of definite integrals known as translation invariance. This means that the value of the integral remains the same even when the function is translated horizontally by a constant amount.
The steps to prove that ∫f(x)dx[b, a] =∫f(x−c)dx [b+c, a+c] are as follows:
1. Rewrite the integral on the left side as ∫f(x)dx[b, a] = ∫f(x)dx[0, a] - ∫f(x)dx[0, b]
2. Use the substitution u = x-c in the first integral on the right side, giving ∫f(x-c)dx[0, a]
3. Use the substitution u = x-c in the second integral on the right side, giving ∫f(x-c)dx[0, b]
4. Rewrite the integral on the right side as ∫f(x-c)dx[b+c, a+c] = ∫f(x-c)dx[a+c, a+c] - ∫f(x-c)dx[b+c, b+c]
5. Simplify the integrals on the right side to get ∫f(x)dx[b, a] = ∫f(x-c)dx[b+c, a+c], proving the original statement.
Yes, this property of definite integrals can be extended to any constant c, not just horizontal translation. This is known as the general translation property of definite integrals, which states that the value of the integral remains the same even when the function is translated horizontally or vertically by a constant amount.
