Find the area of the region under the curve y=x+2 over the interval [1,4].
(1/2)x^2+2x at 4 - at 1
= 8+8 - 1/2 -2
= 13.5
assuming by under you mean below the line and above the x axis
To find the area of the region under the curve y=x+2 over the interval [1,4], we can use definite integration.
The definite integral of a function represents the area under the curve between the given limits. In this case, we want to find the area between the curve y=x+2 and the x-axis from x=1 to x=4.
The first step is to find the antiderivative (also known as the indefinite integral) of the function. By integrating y=x+2, we get:
∫(x+2) dx = (x^2/2) + 2x + C
Next, we evaluate the antiderivative at the upper and lower limits of integration.
Let's substitute our limits into the antiderivative:
[(4^2/2) + 2(4) + C] - [(1^2/2) + 2(1) + C]
Simplifying, we get:
[(16/2) + 8 + C] - [(1/2) + 2 + C]
= (8 + 8 + C) - (1/2 + 2 + C)
= 16 + C - 5/2 - C
The C terms cancel out, leaving us with:
= 16 - 5/2
= 27/2
Therefore, the area under the curve y=x+2 over the interval [1,4] is 27/2 square units.