Question:
The area of the region lying in the first quadrant and bounded by the curve \(y = 4x^2\), the Y-axis, and the lines \(y = 2\), \(y = 4\) is...
The area of the region lying in the first quadrant and bounded by the curve \(y = 4x^2\), the Y-axis, and the lines \(y = 2\), \(y = 4\) is...
Show Hint
Integrate x as a function of y between y = 2 and y = 4.
Updated On: Oct 1, 2026
- \(\frac{1}{3}[8+2\sqrt{2}]\) sq. units
- \(\frac{1}{3}[8-2\sqrt{2}]\) sq. units
- \(\frac{1}{2}[8-2\sqrt{2}]\) sq. units
- \(\frac{1}{2}[8+2\sqrt{2}]\) sq. units
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
The region is bounded by the Y-axis, the curve and two horizontal lines, so we integrate along the Y-axis: \(A = \int_{y_1}^{y_2} x\,dy\).
Step 2: Express x in terms of y:
From \(y = 4x^2\), in the first quadrant \(x = \frac{\sqrt y}{2}\).
Step 3: Integrate:
\[ A = \int_2^4 \frac{\sqrt y}{2}\,dy = \frac12\cdot\frac23\left[y^{3/2}\right]_2^4 = \frac13\left(8 - 2\sqrt2\right) \]
because \(4^{3/2} = 8\) and \(2^{3/2} = 2\sqrt2\).
Step 4: Why the other options are wrong.
Option (A) has \(8 + 2\sqrt2\), a sign slip when subtracting the lower limit. Options (C) and (D) have the factor \(\frac12\), which comes from forgetting the \(\frac23\) in the antiderivative.
Final Answer:
The area is \(\frac13[8 - 2\sqrt2]\) sq. units, option (B). \[ \boxed{\frac{1}{3}\left[8-2\sqrt{2}\right]\text{ sq. units}} \]
The region is bounded by the Y-axis, the curve and two horizontal lines, so we integrate along the Y-axis: \(A = \int_{y_1}^{y_2} x\,dy\).
Step 2: Express x in terms of y:
From \(y = 4x^2\), in the first quadrant \(x = \frac{\sqrt y}{2}\).
Step 3: Integrate:
\[ A = \int_2^4 \frac{\sqrt y}{2}\,dy = \frac12\cdot\frac23\left[y^{3/2}\right]_2^4 = \frac13\left(8 - 2\sqrt2\right) \]
because \(4^{3/2} = 8\) and \(2^{3/2} = 2\sqrt2\).
Step 4: Why the other options are wrong.
Option (A) has \(8 + 2\sqrt2\), a sign slip when subtracting the lower limit. Options (C) and (D) have the factor \(\frac12\), which comes from forgetting the \(\frac23\) in the antiderivative.
Final Answer:
The area is \(\frac13[8 - 2\sqrt2]\) sq. units, option (B). \[ \boxed{\frac{1}{3}\left[8-2\sqrt{2}\right]\text{ sq. units}} \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Area under Simple Curves Questions
Area of the Region bounded by the curve y=√49-x2 and x-axis is .
- Calculate the area of the region bounded by the curve \( y^2 = 4x \) and the line \( x = 3 \).
- Find the area of the region bounded by the curve \(y^2 = 8x\) and the line \(x = 2\).
- The area bounded by the curve $y = |x - 2|$, $x = 1$, $x = 3$ and X-axis is
- The area of the region bounded by the parabola ( y^2 = 27x ) and the line ( x = 1 ) is ________ sq.units.
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions