Find the area of the region bounded by the curves y=sqrtx,y=x^(2), and the lines x=0 and x=1

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Find the area of the region bounded by the curves  y=sqrtx,y=x^(2), and the lines  x=0 and  x=1

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Find Intersection Points: First, we need to find the points of intersection between the curves y=sqrt(x) and y=x^2 within the interval [0, 1]. To do this, we set the two equations equal to each other: sqrt(x) = x^2. Squaring both sides gives us x = x^4. This simplifies to x^4 - x = 0, which factors to x(x^3 - 1) = 0. The solutions to this equation are x = 0 and x = 1, since x^3 - 1 = 0 when x = 1.Determine Upper Curve: Next, we need to determine which curve is above the other between x = 0 and x = 1. For 0 < x < 1, we know that sqrt(x) > x^2 because the square root function grows faster than the square function in this interval. Therefore, y=sqrt(x) is the upper curve and y=x^2 is the lower curve.Set Up Integral: Now we can set up the integral to find the area between the two curves from x = 0 to x = 1. The area A is given by the integral from 0 to 1 of (upper curve - lower curve) dx. So, A = ∫ from 0 to 1 of (sqrt(x) - x^2) dx.Calculate Antiderivatives: We calculate the integral separately for each function. First, for sqrt(x), the antiderivative is (2/3)x^(3/2). Second, for x^2, the antiderivative is (1/3)x^3. So, the integral becomes A = [(2/3)x^(3/2) - (1/3)x^3] from 0 to 1.Evaluate Integral: We evaluate the antiderivatives at the upper and lower limits of the integral. At x = 1, we have (2/3)(1)^(3/2) - (1/3)(1)^3 = (2/3) - (1/3) = 1/3. At x = 0, both terms are zero because any nonzero number raised to a positive power and then multiplied by zero is zero. So, the area A is (1/3) - 0 = 1/3.

Find the area of the region bounded by the curves $y=\sqrt{x}$ , $y=x^{2}$ , and the lines $x=0$ and $x=1$

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Find the area of the region bounded by the curves $y=\sqrt{x}$ , $y=x^{2}$ , and the lines $x=0$ and $x=1$