The area between the curves y=x^2 and y=2x-1 over the interval [0,2] is 8/3 square units.

<p>To find the area between the curves y = x^2 and y = 2x - 1 over the interval [0,2], we need to first find the points of intersection between the two curves.</p><p>Setting the two equations equal to each other, we get:</p><p>x^2 = 2x - 1</p><p>Rearranging and factoring, we get:</p><p>x^2 - 2x + 1 = 0</p><p>(x - 1)^2 = 0</p><p>So the two curves intersect at x = 1.</p><p>Next, we need to find which curve is above the other over the interval [0,2]. We can do this by plugging in values for x and seeing which equation gives a higher y-value.</p><p>When x = 0, y = 0^2 = 0 for the curve y = x^2 and y = 2(0) - 1 = -1 for the curve y = 2x - 1. Therefore, y = x^2 is above y = 2x - 1 when x = 0.</p><p>When x = 2, y = 2^2 = 4 for the curve y = x^2 and y = 2(2) - 1 = 3 for the curve y = 2x - 1. Therefore, y = 2x - 1 is above y = x^2 when x = 2.</p><p>Therefore, the area between the curves over the interval [0,2] is given by:</p><p>A = ∫[0,1] (2x - 1 - x^2) dx + ∫[1,2] (x^2 - 2x + 1) dx</p><p>Evaluating the integrals, we get:</p><p>A = [x^2 - x^3/3] from 0 to 1 + [x^3/3 - x^2 + x] from 1 to 2</p><p>A = (1/3) - (1/3) + (8/3) - 4 + 2 - (1/3) + (1/3) - 2 + 1</p><p>A = 8/3</p><p>Therefore, the area between the curves y = x^2 and y = 2x - 1 over the interval [0,2] is 8/3 square units.</p>

<p>To find the area between the curves y = x^2 and y = 2x - 1 over the interval [0, 2], we need to determine the points of intersection of the two curves and then calculate the definite integral of their difference over that interval.</p><p>First, let's find the points of intersection by setting the two equations equal to each other:</p><p>x^2 = 2x - 1</p><p>Rearranging the equation:</p><p>x^2 - 2x + 1 = 0</p><p>This equation can be factored as:</p><p>(x - 1)^2 = 0</p><p>Therefore, the two curves intersect at x = 1.</p><p>Next, we need to determine which curve is above the other over the interval [0, 2]. To do this, we can evaluate the y-values of each curve at the interval endpoints.</p><p>For the curve y = x^2:</p><ul><li>At x = 0, y = (0)^2 = 0</li><li>At x = 2, y = (2)^2 = 4</li></ul><p>For the curve y = 2x - 1:</p><ul><li>At x = 0, y = 2(0) - 1 = -1</li><li>At x = 2, y = 2(2) - 1 = 3</li></ul><p>From this analysis, we can see that the curve y = x^2 lies below the curve y = 2x - 1 over the interval [0, 2].</p><p>Now, we can calculate the area between the curves by subtracting the equation of the curve below from the equation of the curve above and integrating over the interval [0, 2]:</p><p>Area = ∫[0, 2] [(2x - 1) - x^2] dx</p><p>Simplifying the integrand:</p><p>Area = ∫[0, 2] (2x - 1 - x^2) dx</p><p>Evaluating the integral:</p><p>Area = [x^2 - (1/3)x^3] evaluated from 0 to 2</p><p>Area = [(2)^2 - (1/3)(2)^3] - [(0)^2 - (1/3)(0)^3]</p><p>Area = [4 - (8/3)] - [0 - 0]</p><p>Area = (12/3 - 8/3)</p><p>Area = 4/3</p><p>Therefore, the area between the curves y = x^2 and y = 2x - 1 over the interval [0, 2] is 4/3 square units.</p>