Application of the Integrals MCQ

Ans. The correct answer is d. 1π sq. units

Explanation: Given

The equation of the circle is x² + y² = 1

We have, the center of the circle as (0, 0) and the radius of the circle as 1.

⇒ y² = 1 - x²

⇒ y = √(1 - x²)

The area of the region is given by

A = 4[₀∫¹√(1 - x²) dx]

⇒ A = 4 x (\(\frac{1}{2}\)) x (\(\frac{\pi}{2}\)) = π sq. units

Ans. The correct answer is c. 3π sq. units

Explanation: For the given equation y = √(9 − x²), the curve intersects the x-axis when x = -3 and x = 3

Therefore, the required area is given by

A = _-3∫³ √(9 − x²) dx

On solving, we get

A = 3π sq. units

Ans. The correct answer is a. \(\frac{8}{3} \;a^2\)

Explanation: The equation of the parabola is y² = 4ax

The equation of latus rectum is x = a

Required Area is, A = 2 ₀∫^a y dx = 2 ₀∫^a 2√a √x dx

On solving, we get

A = \(\frac{8}{3} \;a^2\)

Ans. The correct answer is d. 4

Explanation: The required area is given by

A = ₀∫^π sin x dx + _π∫^2π sin x dx 

⇒ A = [- cos x]₀^π + [-cos x]^2π_π

⇒ A = 4 sq units

Ans. The correct answer is b. \(\frac{8}{3}\)

Explanation: The intersection points are (0, 0) and (4, 4), for the given curves y2 = 4x and y = x

The area bounded by the curve is given by

A = ₀∫⁴ [√(4x) - x] dx

⇒ A = 2[ ₀∫⁴√x dx – ₀∫⁴ x dx]

On solving, we get

A = \(\frac{8}{3}\)

Ans. The correct answer is c. 2 sq. units

Explanation: Given, y = cos x and also given that x = 0 and x = π

The required area, A =₀∫^π |cos x| dx

The above equation can also be written as,

A = 2[₀∫^π/2 cos x dx]

⇒ A = 2[sin x]₀^π/2

On solving, we get

A = 2 sq. units

Ans. The correct answer is d. 15/4 sq. units

Explanation: The required area is given by 

A = ₁∫² x³ dx

⇒ A = [x⁴/4]₁²

Now, apply the limits, we get

A = [(2⁴/4) – (1/4)]

⇒ A = 15/4

Ans. The correct answer is b. 6 sq. units

Explanation: The required area is given by 

A =_-1∫¹(2y + 3)dy

⇒ A = [(2y²/2) + 3y]^-1₁

On solving, we get

A = 1+3 -1+ 3

⇒ A = 6 sq. units.

Ans. The correct answer is b. 45/4

Explanation: The required area is given by

A = ₁∫⁸ y^1/3 dy

⇒ A = [(y^4/3)/(4/3)]₁⁸

On solving, we get

A = (3/4)(16-1)

A = (3/4)(15) = 45/4.

Ans. The correct answer is b. 9/8 sq. units

Explanation: For the given curves x² = 4y and x = 4y-2, the points of intersection are x = -1 and x = 2

Therefore the required area is given by

A = _-1∫² [(x + 2)/4)- (x²/4)] dx

On solving, we get

A = (1/4)[(10/3)-(-7/6)]

A = (1/4)(9/2)

⇒ A = 9/8 sq. units

Ans. The correct answer is c. 1

Explanation: The equation of the curve is given as y = loge x

At, x = 1, we get 

y = log_e (1) = 0

At, x = e, we get

y = log_e (e) = 1

Therefore, the required area is given by

A = ₁∫^e log_e x dx

Using integration by parts,

A = [x log_e x – x]₁^e

On solving, we get

A = [e - e - 0 + 1] = 1

Ans. The correct answer is b. 49π/2

Explanation: For the given equation y = √(49 − x²), the curve intersects the x-axis when x = -7 and x = 7

Therefore, the required area is given by

A = _-7∫⁷ √(49 − x²) dx

On solving, we get

A = 49π/2 sq. units

Ans. The correct answer is c. 8π sq. units

Explanation: For the given equation y = √(16 − x²), the curve intersects the x-axis when x = -4 and x = 4

Therefore, the required area is given by

A = _-4∫⁴ √(16 − x²) dx

On solving, we get

A = 8π sq. units

Ans. The correct answer is d. \(\frac{9}{2}\)

Explanation: For the given equation y = x² + 1 and the straight line x + y = 3, the curve intersects the x-axis when x = -2 and x = 1.

Therefore, the required area is given by

A = _-2∫¹ [(3 - x) - (x² + 1)] dx

On solving, we get

A = \(\frac{9}{2}\)

Ans. The correct answer is a. 6π sq. units

Explanation: The given equation is x²/4 + y²/9 = 1

The standard form of ellipse is given by

x²/a² + y²/b² = 1

By comparing we get

a = 2 and b = 3

⇒ y = 3 √(1 - x²/4)

The area bounded by the ellipse is given by

A = 4 x ₀∫² 3 √(1 - x²/4) dx

On solving the above, we get

A = 6π sq. units