Calculate the area of the figure bounded by the lines: a straight line passing through the points (1; 0) and (0; –3); f (x) = – x ^ 2 + 4x-3
Let us calculate the equation of the straight line passing through these points.
Because the general equation of the straight line is y (x) = k * x + b, then we get the system of equations:
k + b = 0 and b = -3, whence we find k = 3.
Therefore, the desired equation of the straight line is y (x) = 3 * x – 3.
We find the abscissas of the intersection points of the straight line and the parabola, we get:
-x² + 4 * x – 3 = 3 * x – 3,
x² – x = 0,
x * (x – 1) = 0, whence x = 0 and x = 1.
Having completed the schematic construction of both graphs, we get that the parabola is located above the straight line, so the required area is:
s = integral (0 to 1) (-x² + 4 * x – 3 – 3 * x + 3) dx = integral (0 to 1) (-x² + x) = 1/2 – 1/3 = 1 / 3 units ².
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