Students can Download Basic Maths Question Bank Chapter 21 Definite Integral and its Applications to Areas Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Basic Maths Question Bank Chapter 21 Definite Integral and its Applications to Areas

### 2nd PUC Basic Maths Definite Integral and its Applications to Areas One Mark Questions and Answers

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Question 7.

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Question 8.

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Question 9.

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Question 10.

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Question 11.

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Question 12.

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Question 15.

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Question 16.

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Question 19.

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Question 20.

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Question 21.

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Question 22.

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= 3 + log(3 + 2) – 0 +1og(0 + 2) = 3 + log 5 – log 2 = 3 + log \(\frac { 5 }{ 2 }\)

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Question 25.

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Question 27.

If marginal cost = x^{2} – x – 2. Find the total cost obtained from an output of 40 units

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Question 28.

If the marginal revenue is given by f^{1}(R) = 15 – \(\frac{x}{15}\) find the Total revenue obtoined from an output of 30 units

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Question 29.

If the marginal cost of a firm is f^{1}(x) 10 + 6x – 6x^{2} where xis the output find the total cost among that the fixed cost is ₹ 125.

Answer:

TC = ∫MC dx

∫10 + 6x – 6x^{2} dx

= 10x + \(\frac{6 x^{2}}{2}-\frac{6 x^{3}}{3}\) + C given C = 125 = fixed cost

= 10x + 3x^{2} – 2x^{3} + 125

Question 30.

If the M R is f^{1}(x) = 20 Find the total revenue & average revenue obtained from an output of 30 units

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Question 31.

Compute the total cost for the marginal cost function f^{1}(x) = 3x^{2} – 6x + 12 assuming that the fixed cost is ₹500 Also find Average cost

Answer:

T C ∫MC dx

Where fixed cost C 500 Rs.

= ∫(3^{2}6x + 12)dx = 3\(\frac{x^{3}}{3}\) – 6x\(\frac{x^{2}}{2}\) + 12x + C = x^{3} – 3x^{2} + 12x + 500

Ave cost = \(\frac{\mathrm{TC}}{x}\)

\(\frac{x^{3}-3 x^{2}+12 x+500}{x}\)

x^{2} – 3x + 12 + \(\frac{500}{x}\)

Question 32.

If the M R = 16 – x^{2} find the maximun total Revenue Also find the total & the average reveues also write the demand function

Answer:

TR ∫MR dx

= ∫(16 – x^{2})dz

l6x – \(\frac{x^{3}}{3}\) + C when output = 0,

TR = 16x – \(\frac{x^{3}}{3}\) Total revenue = 0 ∴ C = 0

Average revenue = \(\frac{\mathrm{TR}}{x}=\frac{16 x-\frac{x^{3}}{3}}{x}=16-\frac{x^{2}}{3}\)

Demand function = 16 – \(\frac{x^{3}}{3}\)

Question 33.

The marginal cost function is f^{1}(x) = 3x^{2} + 2x + 1 wherex is the level of output find the total cost, ave cost, total variable cost, average variable cost given that fixed cost 70

Answer:

TC ∫MCdx

= ∫(3x^{2} + 2x + 1)

= 3 \(\frac{x^{3}}{3}\) + 2\(\frac{x^{2}}{2}\) + x + C given fixed cost C = 70

= x^{3} + x^{2} + x + 70

Total variable cost = 2x^{3} + x^{2} + x

Average cost = \(\frac{\text { Total cost }}{x}\)

= \(\frac{x^{3}+x^{3}+x+70}{x}\) = x^{3} + x + 1 + \(\frac{70}{x}\)

Average variable cost = 2x^{3} + x + 1

Question 34.

Find the area bounded by the curve y = x^{2}, x – axis and the ordinates are x = 0, x = 1

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Question 35.

Find the area bounded by the curve y^{2} = 8x, x – axis & the lines x = 0, x = 2

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Question 36.

Find the area bounded by the curve x^{2} = 8y, y – axis and abscissas y = 3. y = 6 ∵ x = \(\sqrt{8 y}=\sqrt{8} \cdot \sqrt{y}\)

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### 2nd PUC Basic Maths Definite Integral and its Applications to Areas Five Marks Questions and Answers

Question 1.

Find the area between the curves y^{2} = 4x & x^{2} = 4y

Answer:

Given y^{2} = 4x but y = \(\frac{x^{2}}{4}\)

\(\left(\frac{x^{2}}{4}\right)^{2}\) = 4x

\(\frac{x^{4}}{16}\) = 4x ⇒ x^{4} – 64x = 0 ⇒ x(x^{3} – 4^{3}) = 0

⇒ x = 0 or x = 4

Required Area A = |A_{1} – A_{2}|

Question 2.

Find the area bounded by the curve y^{2} = 5x and the line y = x

Answer:

y^{2} = 5x and y = x

⇒ x^{2} = 5x

⇒ x^{2} = 5x

⇒ x^{2} – 5x = 0

⇒ x(x – 5) = 0 ⇒ x = 0 or x = 5 are the limits

Required Area A = |A_{1} – A_{2}|

Question 3.

Find the area between the curves x^{2} = 5y & y = 2x.

Answer:

Given x^{2} = 5y & y = 2x

∴ x^{2} = 5 (2x)

x^{2} = 10x ⇒ x(x – 10) = 0 ⇒ x = 0 or x = 10

RequiredArea = A = |A_{1} – A_{2}|

Question 4.

Find the area bounded by the parabolas x^{2} = y & y^{2} = 8x

Answer:

Given y^{2} = 8 & y = x^{2}

∴ (x^{2})^{2} = 8x

x^{4} – 8x = 0

⇒ x(x^{3} – 2^{3}) =0 ⇒ x = 0 or x = 2

RequiredArea A = |A_{1} – A_{2}|

Question 5.

Find the ares exciosed between the parabola y^{2} = x and the Iine x + y = 2

Answer:

Given y = 2 – x & y^{2} = x

⇒ (2 – x)^{2} = x

4 + x^{2} – 4x – x = 0

x^{2} – 5x + 4 = 0

x = 4 or 1 (both are +v)

If x = 4, y = 2 – x = 2 – 4 = 2

If x = 1, y = 2 – x = 2 – 1 = 1

Question 6.

Find the area between the parabolas y^{2} = 4x & the line y = x

Answer:

Given y^{2} = 4x & y = x

⇒ x^{2} = 4x

⇒ x(x – 4) = 0 ⇒ x = 0 or x = 4

Required area A = A_{1} – A_{2}

Question 7.

Find the area between the cures y = 8 – x^{2} & y = x^{2}

Answer:

Given y = 8 – x^{2} & y = x^{2}

x^{2} = 8 – x^{2}

2x^{2} = 8 ⇒ x^{2} = 4 ⇒ x = ±2

Question 8.

Find the area bounded by the parabola y^{2} = 4ax & its latus rectum

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Required Area = 2 Area OAB

Question 9.

Find the arca exclosed between the parabola x^{2} = 4y and the line x = 4y – 2

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Question 10.

Find the area exciosed between y = 11x – 24 – x^{2} & the line y = x

Answer:

Given y = 11x – 24 – x^{2} & y = x

⇒ x = 11x – 24 – x^{2} ⇒ x^{2} – 10x + 24 = 0

⇒ (x – 4)(x – 6)

⇒ x = 4 or 6