The area (in sq. units) of the region A = x , y ∈ R × R 0 ≤ x ≤ 3 , 0 ≤ y ≤ 4 , y ≤ x 2 + 3 x is

The area of the region O C D is = ∫ 0 1 4 - x 2 + 3 x d x = 4 x - x 3 3 - 3 x 2 2 0 1 = 4 - 1 3 - 3 2 = 13 6 So, the area of the given region O A B C is = Area of O A B D - Area of O C D = 3 × 4 - 13 6 = 59 6

Start Free Mock LoginAlready registered? Login here

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesHard2 minPYQ_2019

MathematicsHardsingle choice

The area (in sq. units) of the region $A = \{(x, y) \in R \times R| 0 \leq x \leq 3, 0 \leq y \leq 4, y \leq x^{2} + 3 x\}$ is

Options:

Answer:

Solution:

The area of the region $O C D$ is
$= \int_{0}^{1} [4 - (x^{2} + 3 x)] d x = {[4 x - \frac{x^{3}}{3} - \frac{3 x^{2}}{2}]}_{0}^{1} = 4 - \frac{1}{3} - \frac{3}{2} = \frac{13}{6}$
So, the area of the given region $O A B C$ is
$=$ Area of $O A B D$ $-$ Area of $O C D$
$= 3 \times 4 - \frac{13}{6}$
$= \frac{59}{6}$

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2019

Doubts & Discussion

Loading discussions...