Find the area of the region bounded by the following curves, the $X -$ axis and the given lines.A
$y = \sqrt{16 x + x^{2}}, x = 0, x = 4$

Open in App

Solution

Area $= \int_{0}^{4} \sqrt{16 x + x^{2}} = \int_{0}^{4} \sqrt{{(x + 8)}^{2} - 64}$
Area $= {∣ ∣ ∣ \frac{(x + 8)}{2} \sqrt{{(x + 8)}^{2} - 64} - \frac{64}{2} l n ∣ ∣ ∣ (x + 8) + \sqrt{{(x + 8)}^{2} - 64} ∣ ∣ ∣ ∣ ∣ ∣}_{0}^{4}$
Area $= 6 \sqrt{80} - 32 l n (12 + \sqrt{80}) + 32 l n (8)$
Area $= 6 \sqrt{80} - 32 l n ∣ ∣ ∣ \frac{12 + \sqrt{80}}{8} ∣ ∣ ∣$ sq units

Suggest Corrections

Similar questions

X -

y = x^{4}, x = 1, x = 5

y = 4 - x^{2}

x = 0

x = 2

y = \sqrt{4 - x^{2}}, x^{2} + y^{2} - 4 x = 0

y = \sqrt{16 - x^{2}}, x = 0

x = 4

y = 1 + \frac{8}{x^{2}}

Join BYJU'S Learning Program