Find the area bounded under the curve $y = 3 x^{2} + 6 x + 7$ and the X-axis with the ordinates at $x = 5$ and $x = 10$ .

Open in App

Solution

So to find the area under the curve $y = 3 x^{2} + 6 x - 7$ and with y-axis as the ordinates at $x = 5$ and $x = 10$ .
$A t x = 5 y = 11 A t x = 10 y = 367$
Area required = dark shaded area
$= \int_{112}^{367} x d y = \int_{112}^{367} \frac{1}{\sqrt{3}} \sqrt{y + 2} - 1 d x = {[\frac{1}{\sqrt{3}} \frac{(y + 2)^{3 / 2}}{3 / 2} - y]}_{112}^{367} = {[\frac{2}{3 \sqrt{3}} (y + 2)^{3 / 2} - y]}_{112}^{3 / 2} = \frac{2}{3 \sqrt{3}} [{(369)}^{3 / 2} - {(11 y)}^{3 / 2}] - [369 - 11 y] = \frac{2}{3 \sqrt{3}} [{27 (41)}^{3 / 2} - {3 \sqrt{3} (38)}^{3 / 2}] - 255 = 2 [3 \sqrt{3} {(41)}^{3 / 2} - {(38)}^{3 / 2}] - 255 s q . u n i t s$

With x-axis and $x = 5, x = 10$

Required area = light shaded area
$= \int_{5}^{10} (3 x^{2} + 6 x + 7) d x = {[x^{2} + 3 x^{2} + 7]}_{5}^{2} = 1370 - 235 = 1135 s q . u n i t s$

Hence the correct answer is $1135 s q . u n i t s$

Suggest Corrections

Similar questions

Find the area bounded by the curve y = $3 x^{2}$ +6x+7 and the X-axis with the x=5 and x=10.

y = x | x |

x = - 3

x = 3

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

y = log x

x = 1

x = e

Join BYJU'S Learning Program