Find the area of the region in the first quadrant bounded by the parabola y = 4x² and the lines x = 0, y = 1 and y = 4.

Open in App

Solution

$y = 4 x^{2} represents a parabola, openeing upwards, symmetrical about + ve y - axis and having vertex at O (0, 0) y = 1 is a line parallel to x - axis, cutting parabola at (- \frac{1}{2}, 1) and (\frac{1}{2}, 1) y = 4 is a line parallel to x axis, cutting parabola at (- 1, 1) and (1, 1) x = 0 is the y - axis Consider a horizontal strip of length = |x| and width = d y in the first quadrant Area of approximating rectangle = |x| d y Approximating rectangle moves from y = 1 to y = 4 Area of the curve in the first quadrant enclosed by y = 1 and y = 4 is the required area of the shaded region ∴ Area of the shaded region = \int_{0}^{4} |x| d y \Rightarrow A = \int_{1}^{4} x d y [As, x > 0, |x| = x] \Rightarrow A = \int_{1}^{4} \sqrt{\frac{y}{4}} d y \Rightarrow A = \frac{1}{2} \int_{1}^{4} \sqrt{y} d y \Rightarrow A = \frac{1}{2} {[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}]}_{1}^{4} \Rightarrow A = \frac{1}{2} \times \frac{2}{3} [4^{\frac{3}{2}} - 1^{\frac{3}{2}}] \Rightarrow A = \frac{1}{3} [8 - 1] \Rightarrow A = \frac{7}{3} sq . units ∴ The area enclosed by parabola in the first quadrant and y = 1, y = 4 is \frac{7}{3} s q . u n i t s$

Suggest Corrections

Similar questions

y = 4 x^{2}, x = 0, y = 1

y = 4

y = 4 x^{2}, x = 0, y = 1, y = 4

\frac{7}{a}

y = x^{2}

y = x

y = x^{2}

y = x

y = 4 x^{2}, x = 0

y = 1, y = 4

Join BYJU'S Learning Program