IF ANYONE CAN PLEAAAASEEEE HELP ME
Use the Riemann sum definition of the definite integral to show that
b
3 ¡ò x©÷dx - b©ø -a©ø
a
Here, for simplicity, we assume b > a > 0. Be sure to clearly specify your mesh/grid xi, , and your
choice of sample points Xi, ¡âx, and your
choice of sample points x*
you need to revisit your posts and type in the functions. This copy/paste stuff is getting mangled.
Use ^ for exponents, like x^2 for x squared. I can guess that 3 means ∑.
To show the Riemann sum definition of the definite integral for the given function, we need to construct a partition of the interval [a, b], choose sample points within each subinterval, and take the limit as the mesh size approaches zero.
Let's start by dividing the interval [a, b] into n subintervals. The length of each subinterval, ∆x, can be found by (b - a) / n.
The mesh/grid points xi are defined as follows:
x₀ = a
x₁ = a + ∆x
x₂ = a + 2∆x
...
xᵢ = a + i∆x
...
xₙ = a + n∆x
Now, we need to choose a sample point x* within each subinterval [xᵢ₋₁, xᵢ]. We can choose any value within the subinterval, but for simplicity, we will choose the right endpoint xᵢ for each subinterval.
Therefore, the sample points Xi can be defined as:
X₁ = x₁
X₂ = x₂
...
Xᵢ = xᵢ
...
Xₙ = xₙ
Using these mesh/grid points and sample points, we can express the Riemann sum for the given definite integral as follows:
∑ f(x*)∆x
To calculate the Riemann sum, we need to substitute the expression for ∆x, sample points Xᵢ, and the function f(x) = x³ into the sum:
∑ (Xᵢ)³∆x
Now, we can simplify the expression:
∑ (Xᵢ)³∆x = ∑ (a + i∆x)³ * ∆x
Expanding the binomial:
∑ (a + i∆x)³ * ∆x = ∑ (a³ + 3a²i∆x + 3ai²(∆x)² + (i∆x)³) * ∆x
Now, we can distribute the ∆x and collect similar terms within the summation:
∑ (a³∆x + 3a²i(∆x)² + 3ai²(∆x)³ + i³(∆x)⁴)
The summation represents the sum of the terms for i = 1 to n.
Now, taking the limit as n approaches infinity (or as ∆x approaches zero), the sum becomes an integral:
lim(n→∞) ∑ (a³∆x + 3a²i(∆x)² + 3ai²(∆x)³ + i³(∆x)⁴)
= ∫[a,b] (a³dx + 3a²x(dx)² + 3ax²(dx)³ + x³(dx)⁴)
Integrating each term, we get:
= [a³x + a²x² + ax³/3 + x⁴/4] evaluated from a to b
Evaluating the definite integral, we get:
= (b³a + a²b² + a * b³/3 + b⁴/4) - (a³a + a²a² + a * a³/3 + a⁴/4)
= b⁴ - a⁴ - a⁴/3 + b⁴/3 - a³ - a³/3 + b³/3 - a²/3 - a² - a²/3 + b²/3
= b⁴ - a⁴ - (a⁴/3 - a³ + b³/3 - a² - a²/3) + (b²/3 - a²/3)
= b⁴ - a⁴ - (a³ - a⁴/3 + b³/3 - a²/3 - a²/3) + (b²/3 - a²/3)
= b⁴ - a⁴ - (a³ - a⁴/3 + a⁴/3 + b³/3 - a²/3 - a²/3) + (b²/3 - a²/3)
= b⁴ - a⁴ - (a³ + b³/3 - a²/3 - a²/3) + (b²/3 - a²/3)
= b⁴ - a⁴ - a³ - b³/3 + a²/3 + a²/3 + b²/3 - a²/3
= b⁴ - a⁴ - a³ - b³/3 + a² + b²/3
Hence, we have shown that the Riemann sum definition of the definite integral ∫(a to b) x³dx is given by:
b
∫ x³dx = b⁴ - a⁴ - a³ - b³/3 + a² + b²/3
a