Explanation:
we have 2 choices :
since f(x) is a line creating a quadrilateral with the x-axis and the coordinates of the interval end points, we could do a geometric solution (area of the given quadrilateral).
or we could calculate the integral of the function over the given interval.
did you learn about integrals already ?
since I am not sure, I will solve this in both ways.
first the geometric solution.
the quadrilateral has the corners (vertexes) at
(3, 0)
(4, 0)
(3, f(3)) = (3, 4×3 + 5) = (3, 17)
(4, f(4)) = (4, 4×4 + 5) = (4, 21)
since the connecting lines from the points on the x-axis to the line are parallel, we could simply rotate the shape by 90° and consider this a trapezoid.
the area of a trapezoid is
height × (baseline + top line) / 2
height = distance between the points on the x-axis
= 4 - 3 = 1
the baseline and the top line are the y-coordinates of the points on the line : 17 and 21.
so, the area is
so, the area is1 × (17 + 21) / 2 = 38/2 = 19 units²
now for the integral solution. if you understand how integrating works, it is very simple. but I will not explain it here. if you have learned it and need further help, please let me know. if you have not learned it yet, please ignore this and just use the first solution above.
the integral of f(x) is
F(x) = 2x² + 5x (+ C)
the area of f(x) over the interval is then
the area of f(x) over the interval is thenF(4) - F(3) = 2×4² + 5×4 (+ C) - 2×3² - 5×3 - (+ C) =
the area of f(x) over the interval is thenF(4) - F(3) = 2×4² + 5×4 (+ C) - 2×3² - 5×3 - (+ C) = = 32 + 20 - 18 - 15 = 52 - 33 = 19 units²
as expected both results are identical.