the integral sign where 4 is on top and 2 is at the bottom (x/2 + 3)dx and im trying to find the areas to evaluate the intrgral
Compute the indefinite integral of(x/2) + 3.
It equals [x^2/4] + 3x
Subtract the value of that function at x=2 from the value at x=4. That will be the area under the curve.
I got 7 but the answer in the book says 21
[x^2/4] + 3x at x= 4 is 16
[x^2/4] + 3x at x= 2 is 7
I get 9 for the answer.
Are you sure you copied the integrand correctly? Is it really (x/2) + 3 ?
I cannot explain the discrepancy
To find the area under the curve represented by the integral sign, you can evaluate the integral using the limits of integration. In this case, the limits of integration are 2 at the bottom and 4 at the top.
Here's how you can evaluate the integral step by step:
1. Start by distributing the integrand: (x/2 + 3)dx
2. Integrate each term separately:
∫(x/2)dx + ∫3dx
The integral of x/2 with respect to x is (1/2) * (x^2):
(1/2) * ∫xdx = (1/2) * (x^2/2) = x^2/4
The integral of 3 with respect to x is simply 3x:
∫3dx = 3x
3. Combine the two integrals:
∫(x/2 + 3)dx = x^2/4 + 3x
4. Evaluate the integral between the limits of integration:
Plug in the upper limit (4) and subtract the result with the lower limit (2):
A = ∫[2 to 4] (x/2 + 3)dx
A = [(4^2/4 + 3*4)] - [(2^2/4 + 3*2)]
A = (16/4 + 12) - (4/4 + 6)
A = (4 + 12) - (1 + 6)
A = 16 - 7
A = 9
Therefore, the area under the curve represented by the given integral is 9 square units.