The area of the region bounded by y2 = 4x, the X-axis and the lines x = 1 and x = 4 is _______.

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KollegeApply · Accepted Answer

bb((28)/(3)) sq. units Explanation: The right-handed parabola in this example, y2 = 4x, has its vertex at the origin, and the lines parallel to the y-axis at x = 1 to x = 4 units distance are x = 1, x = 4. Similarly y2 = 4x contains even power of y and is symmetrical about the x-axis. So the required area = Area of ABCD Area of ABCD = int_1^4ydx=int_1^4sqrt(4x)dx It can be written as = 2int_1^4sqrtxdx = 2[(x3/2)/($\frac{3}{2}$)]_1^4 = 2[(2x3/2)/3]_1^4 Substituting the values we get = 2((2(4)$\frac{3}{2}$)/3-(2(1)$\frac{3}{2}$)/3) = 4($\frac{8}{3}$-$\frac{1}{3}$) = 4($\frac{7}{3}$) = $\frac{28}{3}$ sq. units

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