MHT CET · Maths · Area Under Curves

The area of the region bounded by $x^{2} = 4 y, y = 1, y = 4$ and the $y-$ axis lying in the first quadrant is _______ square units.

A $\frac{22}{3}$
B $\frac{28}{3}$
C $30$
D $\frac{21}{4}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{28}{3}$

Step-by-step Solution

Detailed explanation

Required area

A= \int_{1}^{4} x d y = 2 \int_{1}^{4} \sqrt{y} d y

A = 2 {(\frac{y^{\frac{3}{2}}}{\frac{3}{2}})}_{1}^{4}

A = \frac{4}{3} (8 - 1) = \frac{4}{3} (7) = \frac{28}{3}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app

	Column I		Column II
i.	Prostate gland	a.	hyaluronidase
ii.	Acrosome of sperm	b.	Inhibin
Iii.	Corpus luteum	c.	acid phosphatase
iv.	Placenta	d.	Human Chorionic Gonadotropin

The area of the region bounded by x2=4y, y=1, y=4 and the y-axis lying in the first quadrant is _______ square units.

Step-by-step Solution

The area of the region bounded by $x^{2} = 4 y, y = 1, y = 4$ and the $y-$ axis lying in the first quadrant is _______ square units.