CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Application of Integrals

8.2. Area of The Region Bounded by a Curve and a Line.

Q.1. Find the area of the region bounded by the parabola x² = 4y and the line x = 4y – 2.

Solution :

Fig.

We have, x² = 4y ------------------ (1)
And x = 4y – 2 ----------------- (2)
From (1) and (2) we get
x² = x + 2
Or, x² – x – 2 = 0
Or, x = – 1, 2.
Therefore, (1) and (2) intersect at B(2, 1) and C (– 1, 1/4).
Required area = Area of the shaded region
= –1∫² (y2 – y1) dx
= –1∫²[(x + 2)/4 – x²/4] dx
= 1/4 –1∫2(x + 2 – x²) dx
= 1/4 [x² /2 + 2x – x³/3] –12
= 1/4 [(2 + 4 – 8/3) – (1/2 – 2 + 1/3)]
= 1/4 [10/3 – (–7/6)]
= 9/8 sq. units. [Ans.]

Q.2. Find the area of the region bounded by y² = 4x, x = 1, x = 4 and x-axis in the first quadrant.

Solution :

Fig.

The required area = Area of the shaded region
= 1∫⁴ydx
= 1∫⁴√(4x) dx
= 2.[(x3/2)/(3/2)]₁⁴
= 4/3[4^3/2 – 1]
= 28/3 sq. units. [Ans.]

Paper By Mr. M.P.Keshari
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