what is the area in square units of the region under the curve of the function f(x)=x+3, on the interval from x=1 to x=3?
10
12
14
16
18 ?
even without calculus, since this is just a trapezoid with bases 4 and 6, height 2.
5*2 = 10
now, using calculus, integrate under the curve, from 1 to 3
integral[1,3] x+3 dx
= x^2/2 + 3x [1,3]
= (9/2 + 9) - (1/2 + 3) = 10
To find the area under the curve of a function, you can use definite integration. In this case, the function is f(x) = x + 3, and you want to find the area under the curve on the interval from x = 1 to x = 3.
To calculate the area, you need to find the integral of the function over that interval. The integral of f(x) can be found by integrating each term separately. The integral of x is (1/2)x^2, and the integral of 3 is 3x.
So, the integral of f(x) = x + 3 is (1/2)x^2 + 3x.
To find the area under the curve on the interval from x = 1 to x = 3, you subtract the integral at the lower value (x = 1) from the integral at the upper value (x = 3).
(1/2)(3)^2 + 3(3) - [(1/2)(1)^2 + 3(1)] = 9 + 9 - (1/2 + 3) = 18 - (1/2 + 3) = 18 - (7/2) = 18 - 3.5 = 14.5.
Therefore, the area under the curve of the function f(x) = x + 3 on the interval from x = 1 to x = 3 is 14.5 square units.
None of the options (10, 12, 14, 16, 18) are exact matches to 14.5, so the closest option would be 14.