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

Application of Integrals

Q.3. Using integration, find the area bounded by the curve x² = 4y and the straight line x = 4y – 2.

Solution :

Fig.

As in the figure above,
x² = 4y -------------------- (1)
and x = 4y – 2 -------------------- (2)
meet where
(4y – 2)² = 4y
=> 16y² – 16y + 4 = 4y
=> 4y² – 5y + 1 = 0
=> (4y – 1)(y – 1) = 0
=> y = 1, 1/4.
Thus they intersect at A(– 1, 1/4) and B(2, 1).
The required area = –1∫2(y² – y1)dx
= –1∫²[(x + 2)/4 – (x2)/4]dx
= 1/4[(x²)/2 + 2x – (x³)/3]–12
= 1/4[2 + 4 – 8/3 – (1/2 – 2 + 1/3)]
= 9/8 sq. units. [Ans.]

Q.4. Find the area of the region {(x, y) : y ≥ x² and y = |x|}.

Solution :

Fig.

∫ x if x ≥ 0
We have, |x| = | --------------- (1)
∫ – x if x < 0
y = x² ---------------- (2)
We see that (1) and (2) intersect at (0, 0), (1, 1) and (–1, 1).
As, figure is symmetric about y-axis.
Therefore, Area of shaded region = 2[Area of shaded region in XOY plane]
= 2[₀∫¹ x dx – ₀∫¹ x² dx]
= 2{[x²/2]01 – [x3/3]01}
= 2[(1/2 – 0) – (1/3 – 0)]
= 1 – 2/3
= 1/3. sq. unit. [Ans.]

Paper By Mr. M.P.Keshari
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