Integrals MCQs

Ans: (a) 3x²

Explanation: Since derivative of f(x) = f'(x) = 6x = g(x)

Now, if g(x) = 6x, then, the anti-derivative of g(x) = ∫ g(x) = 3x²

Ans: (c)\(\frac{1}{2}\) sin (2x) + C.

Explanation: ∫ \(\frac{d}{dx}\)(f(x)) =∫ cos 2x

• Let us assume that 2x = t
• Thus, x = \(\frac{t}{2}\)
• dx = \(\frac{dt}{2}\)
• The integral thus becomes ∫\(\frac{1}{2}\)(cos t) dt
• = \(\frac{1}{2}\)(sin t) + C = \(\frac{1}{2}\) sin (2x) + C
• Therefore, the integral of cos 2x is \(\frac{1}{2}\) sin (2x) + C.

Ans. (d) \(\frac{1}{4}\) e^4x + C.

Explanation: ∫ \(\frac{d}{dx}\)(f(x)) = ∫ \(\frac{d}{dx}\)( e^4x)

• As we know, the form of integral is ∫ \(\frac{d}{dx}\)( e^ax) = \(\frac{1}{a}\) e^ax + C
• ∫ \(\frac{d}{dx}\)( e^4x) = \(\frac{1}{4}\) e^4x + C
• Therefore, the integral of e^4x is \(\frac{1}{4}\) e^4x + C

Ans: (c) ∫

Explanation: The symbol of integrals is ∫ which indicates the lower and higher limit of a function.

Ans.(a) sin (x² – 3) + C

Explanation: I = ∫3x cos(x² – 3).dx

• Let x² – 3 = t …..(1)
• 2x.dx = dt
• Substituting these values, we have
• I = ∫cos(t).dt
• = sint + c …..(2)
• Substituting the value of 1 in 2, we have
• = sin (x² – 3) + C
• This is the required integration for the given function.

Ans: (c) f(x)

Explanation: As given in the question: (d/dx) f(x) = g(x)

Since we know that the integration is the inverse process of differentiation, the antiderivative of g(x) is f(x).

Ans: (b) \(\frac{16}{3}\)

Explanation: ₀∫² 2x² dx = [2x³/3]⁰₂

Now, apply the limits, we get

₀∫² 2x²dx =(2x2³/3) – 0 = \(\frac{16}{3}\)

Ans: (a) log x + c

Explanation: ∫\((\frac{x^{25}}{x^{26}})\) dx = ∫ x^25-26 dx

•  ∫ x-1 dx
•  ∫ (1/x) dx
•  log x + c

Ans: (a) -cot x – x + C

Explanation: Since it is known that cot² x = cosec 2x – 1

• ∫cot²x dx = ∫ (cosec2x – 1) dx
• -cot x -x + C. [Since, ∫cosec2x dx = – cot x + c]

Ans: (c) 48

Explanation: ₀∫⁴ 6x dx = 6₀∫⁴ x dx = 6[x²/2]⁰₄

Applying the limits formula: ₀∫⁴ 6x dx = 6[(42/2) -0]

• ₀∫⁴ 6x dx = 6[8-0] = 48
• Hence, ₀∫⁴ 6x dx = 48.

Ans: (c) \(- \frac{1}{2}\)

Explanation: Given: ∫ sec²(6– 2x)dx = a tan (6 – 2x) + C

∫ sec²(6 – 2x)dx ={[tan (6-2x)]/-2} + C

∫ sec²(6 – 2x)dx = (\(- \frac{1}{2}\) ) tan (6-2x) + C

Hence, the value of a is - \(\frac{1}{2}\)

Ans: (b) 2 +6x²

Explanation: Since derivative of f(x) = f'(x) = 12x = g(x)

Now, if g(x) = 12x, then, the anti-derivative of g(x) = ∫ g(x) = 2 +6x²

Ans: (a)\(\frac{1}{5}\) sin (5x) + C.

Explanation: ∫ \(\frac{d}{dx}\)(f(x)) =∫ cos 5x

• Let us assume that 5x = t
• Thus, x = \(\frac{t}{5}\)
• dx = \(\frac{dt}{5}\)
• The integral thus becomes ∫\(\frac{1}{5}\)(cos t) dt
• = \(\frac{1}{5}\)(sin t) + C = \(\frac{1}{5}\) sin (5x) + C
• Therefore, the integral of cos 5x is \(\frac{1}{5}\) sin (5x) + C.

Ans. (d) \(\frac{1}{8}\) e^8x + C.

Explanation: ∫ \(\frac{d}{dx}\)(f(x)) = ∫ \(\frac{d}{dx}\)( e^8x)

• As we know, the form of integral is ∫ \(\frac{d}{dx}\)( e^ax) = \(\frac{1}{a}\) e^ax + C
• ∫ \(\frac{d}{dx}\)( e^8x) = \(\frac{1}{4}\) e^8x + C
• Therefore, the integral of e^8x is \(\frac{1}{8}\) e^8x + C

Ans: (c)\(\frac{190}{3}\)

Explanation: Assume that, I = ∫₂³ 10x² dx

• Now,∫₂³ 10x² dx = 10(x³)/3
• 10[(3³)/3 – (2³)/3] = 10[(27/3) – (8/3)]
• 10[(27 – 8)/3] = 190/3
• Thus, ∫₂³ 10x² dx = 190/3

Ans: (c) \(- \frac{1}{2}\)

Explanation: Given: ∫ sec²(10– 2x)dx = a tan (10 – 2x) + C

• ∫ sec²(10 – 2x)dx ={[tan (10-2x)]/-2} + C
• ∫ sec²(10 – 2x)dx = (-½ ) tan (10-2x) + C
• Hence, the value of a is \(- \frac{1}{2}\)

Ans: (b) log x + c

Explanation: ∫\((\frac{x^{27}}{x^{28}})\) dx = ∫ x^27-28 dx

•  ∫ x-1 dx
•  ∫ \((\frac{1}{x})\) dx
•  log x + c