Prove that the integral on the interval [a,b] of x is equal (b^2-a^2)/2
integral a to be (x)dx = (b^2-a^2)/2
using the definition of a Definite Integral. This is the limit of a sum approach.
show steps please!!!
Thank you!!!
To prove the integral of x from a to b is equal to (b^2 - a^2)/2, we can use the definition of a definite integral as the limit of a sum.
First, let's divide the interval [a, b] into n subintervals of equal width, denoted by Δx. The width of each subinterval is given by Δx = (b - a)/n.
Now, let's choose arbitrary points within each subinterval, denoted by xi. These points will be used to evaluate the function x at those intervals.
Next, we can represent the definite integral as the limit of a sum:
∫[a, b] x dx = lim(n→∞) Σ[i=1 to n] f(xi) Δx
where f(xi) is the value of x at xi.
Substituting the expression for Δx, we have:
∫[a, b] x dx = lim(n→∞) Σ[i=1 to n] f(xi) [(b - a)/n]
Now, we can evaluate the right-hand side of the equation. Since we are integrating x from a to b, f(xi) = xi.
Σ[i=1 to n] f(xi) [(b - a)/n] = Σ[i=1 to n] xi [(b - a)/n]
Expanding the summation, we have:
(b - a)/n * (x1 + x2 + x3 + ... + xn)
Using the arithmetic progression formula for the sum of n terms, which is n/2 * (a + l), where n is the number of terms, a is the first term, and l is the last term, we simplify the expression further:
(b - a)/n * (x1 + x2 + x3 + ... + xn) = (b - a)/n * [n/2 * (x1 + xn)]
Since xi represents equally spaced points within the interval, x1 will be equal to a and xn will be equal to b:
(b - a)/n * [n/2 * (x1 + xn)] = (b - a)/n * [n/2 * (a + b)]
Simplifying, we get:
(b - a)/n * [n/2 * (a + b)] = (b - a)/2 * [a + b]
Now, as n approaches infinity, (b - a)/n approaches 0, and we are left with:
(b - a)/2 * [a + b] = (b^2 - a^2)/2
Thus, we have proved that the integral of x from a to b is equal to (b^2 - a^2)/2.