ML Aggarwal Class 12 Maths Solutions Section B Chapter 3 Applications of Integrals MCQs

Utilizing ML Aggarwal Class 12 ISC Solutions Chapter 3 Applications of Integrals MCQs as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section B Chapter 3 Applications of Integrals MCQs

Choose the correct answer from the given four options in questions (1 to 17) :

Question 1.
The area of the region (in square units) bounded by the curve x² = 4y, line x = 2 and x-axis is
(a) 1
(b) 2/3
(c) 4/3
(d) 8/3
Solution:
(b) 2/3

Given eqn. of curve be x² = 4y represents upward parabola with vertex (0, 0).
Now x = 2 meets x² = 4y at y = 1

divide the shaded region into vertical strips with tower end on x-axis and upper end on given parabola.
The strip moves from x = 0 to x = 2.
∴ required area = \(\int_0^2\) y dx
= \(\int_0^2 \frac{x^2}{4}\) dx
= \(\left.\frac{x^3}{12}\right]_0^2\)
= \(\frac{8}{12}=\frac{2}{3}\) sq. units

Question 2.
The area of the region bounded by the curve y = x² and the line y = 16 is
(a) \(\frac{32}{3}\)
(b) \(\frac{256}{3}\)
(c) \(\frac{64}{3}\)
(d) \(\frac{128}{3}\) [NCERT Exemplar]
Solution:
(b) \(\frac{256}{3}\)

Given eqn. of curve be y = x² which represents an upward parabola with vertex (0, 0)
and y = 16 represents a line || to x-axis.

Further, both curves y = x² and line y = 16
intersects when x² = 16
i.e. x = ± 4
i.e. intersects at points (± 4, 16).
∴ Required area of region OCBO = 2 × area of region OABO
= 2 \(\int_0^4\) (16 – x²) dx
= 2 \(\left[16 x-\frac{x^3}{3}\right]_0^1\)
= 2 [64 – \(\frac{64}{3}\)]
= \(\frac{256}{3}\) sq. units

Question 3.
The area (in sq. units) enclosed by the parabola y² = 4ax and its latus-rectum is
(a) \(\frac{4}{3}\) a²
(b) \(\frac{16}{3}\) a²
(c) \(\frac{8}{3}\) a²
(d) none of these
Solution:
(c) \(\frac{8}{3}\) a²

eqn. of given parabola be y² = 4ax
eqn. of latus-rectum be x = a

The end points of latus rectum are (a, 2a) and (a, – 2a)
∴ required area = 2 × Area of region OABA
= 2 \(\int_0^a\) 2√a√x dx
= 4√a \(\left.\frac{x^{3 / 2}}{3 / 2}\right]_0^a\)
= \(\frac{8}{3}\) √a × a^3/2
= \(\frac{8}{3}\) a²

Question 4.
The area bounded by the parabola y² = 8x, the x-axis and the latusrectum is
(a) \(\frac{16}{3}\)
(b) \(\frac{23}{3}\)
(c) \(\frac{32}{3}\)
(d) \(\frac{16 \sqrt{2}}{3}\)
Solution:

Given eqn. of parabola be y² = 8x, which is right handed with vertex (0, 0).
On comparing with y² = 4ax
⇒ 4a = 8
⇒ a = 2

∴ eqn. of latus rectum be x = a
i.e. x = 2
Now the line x = 2 intersects the parabola
y² = 8x when y² = 16
i.e y = ± 4
∴ points of intersection are (2, ± 4).
∴ required area = 2 × area of shaded region OABO divide the shaded region into vertical strips.
each vertical strip has lower end on x-axis and upper end on given parabola.
Clearly the rectangle moves from x = 0 to x = 2.
∴ required area = 2 \(\int_0^2 \sqrt{8} \sqrt{x}\)
= 2\(\left.\sqrt{8} \frac{x^{3 / 2}}{3 / 2}\right]_0^2\)
= \(\frac{4 \sqrt{8}}{3} \times 2^{3 / 2}\)
= \(\frac{32}{3}\) sq.unlts

Question 5.
The area of the region bounded by parabola y² = x and the straight line 2y = x ¡s
(a) \(\frac{4}{3}\) sq units
(b) 1 sq units
(c) \(\frac{2}{3}\) sq units
(d) \(\frac{1}{3}\) sq units
Solution:
(a) \(\frac{4}{3}\) sq units

Given parabola y² = x ………………..(1)
is a right handed parabola with vertex (0, 0)
and line 2y = x …………………..(2)
eqn. (1) and (2) intersects when
y² = 2y
⇒ y (y – 2) = 0
⇒ y = o, 2
∴ From (2) ;
x = 0, 4

∴ points of intersection of both curves are (0, 0) and (4, 2)
Divide the region into vertical strips with lower end on line 2y = x
and upper end on parabola y² = x.
∴ required area = \(\int_0^4\left[\sqrt{x}-\frac{x}{2}\right]\)
= \(\left[\frac{2}{3} x^{3 / 2}-\frac{x^2}{4}\right]_0^4\)
= \(\frac{16}{3}-4\)
= \(\frac{4}{3}\) sq. units

Question 6.
The area of the region (in sq. units) bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is
(a) \(\frac{7}{2}\)
(b) \(\frac{9}{2}\)
(c) \(\frac{11}{2}\)
(d) \(\frac{13}{2}\)
Solution:
(a) \(\frac{7}{2}\)

eqn. of given line be
y = x + 1
⇒ – x + y = 1
which meets coordinate axes are (- 1, 0) and (0, 1)

∴ required area of region ABCDA = \(\int_2^3\) (x + 1) dx
= \(\left.\frac{(x+1)^2}{2}\right]_2^3\)
= \(\frac{1}{2}\) [16 – 9]
= \(\frac{7}{2}\) sq. units

Question 7.
The area of the region bounded by the curve x = 2y + 3 and they lines y = 1 and y = – 1 is
(a) 4 sq. units
(b) \(\frac{3}{2}\) sq. units
(c) 6 sq. units
(d) 8 sq. units
Answer:
(c) 6 sq. units

Equation of given curve be x = 2y + 3 which represents a line meets x-axis at (3, 0) and
y-axis at (0, – \(\frac{3}{2}\))
Thus shaded region R is given by
R = {(x, y) ; 0 ≤ x ≤ 2y + 3 ; – 1 ≤ y ≤ 1}
∴ required area = \(\int_{-1}^1\) (2y + 3) dy
= \(\left.\frac{(2 y+3)^2}{2 \times 2}\right]_{-1}^1\)
= \(\frac{1}{4}\) [25 – 1]
= 6 sq. units.

Question 8.
The area bounded by the curve y = log_e x and x-axis and the straight line x = e is
(a) e sq. units
(b) 1 sq. units
(c) 1 – \(\frac{1}{e}\) sq.units
(d) 1 + \(\frac{1}{e}\) sq.units
Solution:
(b) 1 sq. units

Given eqn. of curve bey = log_e x
When x = e
⇒ y log_e e = 1
∴ required area = \(\int_1^e \log x \cdot 1\) dx
= [x log x – x\(]_1^e\)
= [e log e – e – 1 . log 1 + 1]
= 1 sq. units

Question 9.
If the area above the x-axis, bounded by the curves y = 2^kx and x = 0, and x = 2 is \(\frac{3}{\log _e 2}\) then the value of k is
(a) 1/2
(b) 1
(c) – 1
(d) 2
Solution:
(b) 1

Given eqn. of curve be y = 2^kx
When x = 0
⇒ y = 2⁰ = 1
When x = 2
⇒ y = 2^2k

Question 10.
If the area bounded by the curve y² = 16x and the line y = mx is \(\frac{2}{3}\) sq. units, then the value of m is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4

eqns. of given curves are
y² = 16x ………………(1)
and y = mx …………………(2)
eqn. (1) represents right handed parabola.
both curves (1) and (2) intersects
When m²x² = 16x
⇒ x(m²x – 16) = 0
⇒ x = 0, \(\frac{16}{m^2}\)
∴ y = 0, \(\frac{16}{m}\)

Question 11.
The area (in sq. units) enclosed by the circle x² + y² is
(a) 4π
(b) 2√2π
(c) 2π
(d) 4π²
Solution:
(c) 2π

eqn. of given circle be x² + y² = 2
Required area of circle = 4 × area of region OABO

= 4 × \(\int_0^{\sqrt{2}} \sqrt{2-x^2}\) dx
= 4 \(\left[\frac{x \sqrt{2-x^2}}{2}+\frac{2}{2} \sin ^{-1} \frac{x}{\sqrt{2}}\right]_0^{\sqrt{2}}\)
= 4 [0 + sin^-1 1 – 0 – 0]
= 4 × \(\frac{\pi}{2}\)
= 2π sq. units

Question 12.
The area enclosed by the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 is equal to
(a) π² ab
(b) π ab
(c) π a²b
(d) π ab² [NCERT Exemplar]
Solution:
(b) π ab

Given eqn. of ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
R = \(\left\{\begin{array}{r}
(x, y) ; \frac{-b}{a} \sqrt{a^2-x^2} \leq y \leq \frac{b}{a} \sqrt{a^2-x^2} \\
-a \leq x \leq a
\end{array}\right\}\)

Question 13.
The area (in sq. units) enclosed by the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}\) = 1 is
(a) 20π
(b) 20π²
(c) 16π²
(d) 25π
Solution:
(a) 20π

Given eqn. of ellipse be

Question 14.
The area of the region bounded by the curve y = \(\sqrt{16-x^2}\) and x-axis is
(a) 8π sq. units
(b) 20π sq. units
(c) 16π sq. units
(d) 256π sq. units
Solution:
(a) 8π sq. units

Given equation of curve be
y = \(\sqrt{16-x^2}\)
which represents a circle with centre (0, 0) and radius 4.
∴ shaded region is given by
R = {(x, y) ; 0 ≤ y ≤ \(\sqrt{16-x^2}\) ; – 4 ≤ x ≤ 4}

Thus required area = \(\int_{-4}^4 \sqrt{16-x^2}\) dx
= \(\int_{-4}^4 \sqrt{4^2-x^2}\) dx
= \(\left.\frac{x \sqrt{16-x^2}}{2}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{-4}^4\)
= 0 + 8 sin^-1 1 – 0 – 8 sin^-1 1
= 16 sin^-1 1
= 16 × \(\frac{\pi}{2}\)
= 8π sq. units.

Question 15.
The area bounded by the curve y = sin x and the ordinates x = 0 and x = \(\frac{\pi}{2}\) is
(a) 2 sq. units
(b) 3 sq. units
(c) 4 sq. units
(d) 1 sq. units
Solution:
(d) 1 sq. units

Required area of region = \(\int_0^{\pi / 2}\) y dx
= \(\int_0^{\pi / 2}\) sin x dx
= – cos x\(]_0^{\pi / 2}\)
= – [cos \(\frac{\pi}{2}\) – cos 0]
= – (0 – 1)
= 1 sq. units

Question 16.
The area (in sq. units) bounded by the curve y = cos x and the x-axis between x = 0 and x = π is
(a) 1
(b) 2
(c) 4
(d) 6
Solution:
(b) 2

required area = area of region OABO + area of region AECA

= \(\int_0^{\pi / 2}|\cos x| d x+\int_{\pi / 2}^\pi|\cos x| d x\)
= \(\int_0^{\pi / 2}\) cos x dx – \(\int_{\pi / 2}^\pi\) cos x dx
= \(\left.\sin x]_0^{\pi / 2}-\sin x\right]_{\pi / 2}^\pi\)
= (1 – 0) – (0 – 1)
= 2 sq. units

Question 17.
The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ \(\frac{\pi}{2}\) is
(a) 2 (√2 – 1)
(b) √2 – 1
(c) √2 + 1
(d) √2
Solution:
(b) √2 – 1

eqns. of both curves are;
y = cos x …………….(1)
and y = sin x ………………..(2)

Thus, both curves intersects
when sin x = cos x
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\)
∴ From (1) ;
y = \(\frac{1}{\sqrt{2}}\)
Thus both curves intersects at \(\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)\)
Divide the region into vertical strips with lower end on eqn. (2) and upper end on curve y = cos x.
∴ required area = \(\int_0^{\pi / 4}\) [cos x – sin x] dx
= [sin x + cos x\(]_0^{\pi / 4}\)
= [\(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) – 0 – 1]
= [√2 – 1] sq. units