Area Under Curves

Question

The area bounded by the curves $y = \sqrt{x}, 2y - x + 3 = 0$, and the $x$-axis in the first quadrant is:

RankGuru · Accepted Answer

1. Identify the boundaries:
Curve 1: $y = \sqrt{x}$  (or $x = y^2$).
Curve 2: $x = 2y + 3$.
Boundary 3: $y = 0$ ($x$-axis).

2. Find intersection points:
Solve $x = y^2$ and $x = 2y + 3$.
$y^2 = 2y + 3 \implies y^2 - 2y - 3 = 0 \implies (y-3)(y+1) = 0$.
Since we are in the first quadrant ($y>0$), $y = 3$.
Intersection is at $y=3, x=9$. Point $(9,3)$.

3. Set up the integral with respect to $y$ (Horizontal Strips):
This avoids splitting the area. $y$ goes from 0 to 3.
Right boundary: line $x = 2y + 3$.
Left boundary: curve $x = y^2$.
Area $= \int_0^3 (x_{right} - x_{left}) dy = \int_0^3 ((2y + 3) - y^2) dy$.

4. Evaluate:
$= [y^2 + 3y - y^3/3]_0^3$
$= (9 + 9 - 9) - 0 = 9$.

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Area Under Curves

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More area under curves Practice Questions

The area bounded by the parabolas $y^{2} = 4 x$ and $x^{2} = 4 y$ is:

Area bounded by the curves $y = ln x, y = 0$ , and $x = e$ is:

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Area Under Curves

Select the correct option:

Solution

🔒 Solution Hidden from View

More area under curves Practice Questions

The area bounded by the parabolas y2=4x and x2=4y is:

Area bounded by the curves y=lnx,y=0, and x=e is:

About This Question

Solution

The area bounded by the parabolas $y^{2} = 4 x$ and $x^{2} = 4 y$ is:

Area bounded by the curves $y = ln x, y = 0$ , and $x = e$ is: