Maharashtra Board Class 12 Maths Part 1 Chapter 7 Miscellaneous Solutions

(I) Choose The Correct Alternatives:

Question 1. Area of the region bounded by the curve x² = y, the X-axis and the lines x = 1 and x = 3 is
(a) \(\frac{26}{3}\) sq units
(b) 26 sq units
(c) 26 sq units
(d) 3 sq units
Answer: (a) \(\frac{26}{3}\) sq units
In simple words: To find the area, integrate the curve \(y = x^2\) from \(x = 1\) to \(x = 3\) with respect to x. This represents the total enclosed area.

🎯 Exam Tip: Remember to set up the definite integral correctly using the given curve and boundary lines. Pay attention to the limits of integration.

Question 2. The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is
(a) 28 sq units
(b) 3 sq unit
(c) \(\frac{28}{3}\) sq units
(d) 28 sq units
Answer: (c) \(\frac{28}{3}\) sq units
In simple words: The area is calculated by integrating \(y = \sqrt{4x} = 2\sqrt{x}\) (the upper half of the parabola) from \(x = 1\) to \(x = 4\).

🎯 Exam Tip: When dealing with \(y^2 = 4x\), remember that \(y\) can be positive or negative. For the area bounded by the X-axis, usually, the positive root \(y = \sqrt{4x}\) is used for the region above the X-axis.

Question 3. Area of the region bounded by x² = 16y, y = 1 and y = 4 and the Y-axis, lying in the first quadrant is
(a) 63 sq units
(b) \(\frac{63}{3}\) sq units
(c) \(\frac{56}{3}\) sq units
(d) \(\frac{63}{7}\) sq units
Answer: (c) \(\frac{56}{3}\) sq units
In simple words: To find this area, express \(x\) in terms of \(y\) from \(x^2 = 16y\) as \(x = \sqrt{16y} = 4\sqrt{y}\) and integrate with respect to \(y\) from \(y = 1\) to \(y = 4\).

🎯 Exam Tip: Be careful to integrate with respect to the correct variable (\(x\) or \(y\)) based on the given boundary lines and the axis involved (X-axis or Y-axis).

Question 4. Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is
(a) \(\frac{3142}{5}\) sq units
(b) \(\frac{3124}{5}\) sq units
(c) \(\frac{3142}{3}\) sq units
(d) \(\frac{3124}{3}\) sq units
Answer: (b) \(\frac{3124}{5}\) sq units
In simple words: Integrate the function \(y = x^4\) from the lower limit \(x = 1\) to the upper limit \(x = 5\) to find the area under the curve.

🎯 Exam Tip: Basic power rule for integration, \(\int x^n dx = \frac{x^{n+1}}{n+1}\), is frequently used for such problems. Ensure accurate calculation of the definite integral.

Question 5. Using definite integration area of circle x² + y² = 25 is
(a) 5π sq units
(b) 4π sq units
(c) 25π sq units
(d) 25 sq units
Answer: (c) 25π sq units
In simple words: The standard formula for the area of a circle is \(\pi r^2\). For the equation \(x^2 + y^2 = r^2\), the radius is \(r = 5\), so the area is \(\pi (5)^2 = 25\pi\).

🎯 Exam Tip: Recognize standard geometric shapes and their area formulas. While integration can be used to derive these, knowing the formulas saves time for common shapes like circles or ellipses.

(II) Fill In The Blanks:

Question 1. Area of the region bounded by y = x⁴, x = 1, x = 5 and the X-axis is __________ sq units
Answer: \(\frac{3124}{5}\)
In simple words: This is a direct application of definite integration, calculating \(\int_{1}^{5} x^4 dx\).

🎯 Exam Tip: Practice integration of power functions and evaluation of definite integrals accurately, as small arithmetic errors can lead to incorrect answers.

Question 2. Using definite integration area of the circle x² + y² = 49 is __________
Answer: 49π sq units
In simple words: For a circle \(x^2 + y^2 = r^2\), the radius is \(r = \sqrt{49} = 7\), so the area is \(\pi r^2 = \pi (7)^2 = 49\pi\).

🎯 Exam Tip: Understand that \(r^2\) is the constant on the right side of the circle's equation. Extract \(r\) correctly before applying the area formula.

Question 3. Area of the region bounded by x² = 16y, y = 1, y = 4 and the Y-axis lying in the first quadrant is __________ sq units
Answer: \(\frac{56}{3}\)
In simple words: Integrate \(x = 4\sqrt{y}\) with respect to \(y\) from \(1\) to \(4\) to find the area in the first quadrant.

🎯 Exam Tip: Visualize the region described. Integrating \(x\) with respect to \(y\) is suitable when the region is bounded by the Y-axis and horizontal lines.

Question 4. The area of the region bounded by the curve x² = y, the X-axis and the lines x = 3 and x = 9 is __________
Answer: 234 sq units
In simple words: The area is found by computing the definite integral of \(x^2\) from \(x = 3\) to \(x = 9\).

🎯 Exam Tip: Ensure that the limits of integration correspond to the given x-values. The calculation involves finding \(\int_{3}^{9} x^2 dx\).

Question 5. The area of the region bounded by y² = 4x, the X-axis and the lines x = 1 and x = 4 is __________
Answer: \(\frac{28}{3}\) sq units
In simple words: Integrate \(y = 2\sqrt{x}\) from \(x = 1\) to \(x = 4\) to find the area above the X-axis.

🎯 Exam Tip: Remember to express \(y\) as a function of \(x\) from \(y^2 = 4x\), taking the positive root for the upper region.

(III) State Whether Each Of The Following Is True Or False.

Question 1. The area bounded by the curve x = g(y), Y-axis and bounded between the lines y = c and y = d is given by \(\int_{c}^{d} x dy = \int_{y=c}^{y=d} g(y) dy\)
Answer: True
In simple words: This statement correctly describes how to calculate the area when integrating with respect to \(y\), where the curve is defined as \(x = g(y)\) and boundaries are horizontal lines.

🎯 Exam Tip: Understand the different forms of area calculation using definite integrals, \(\int y dx\) for area with respect to X-axis and \(\int x dy\) for area with respect to Y-axis.

Question 2. The area bounded by two curves y = f(x), y = g(x) and X-axis is \(\left| \int_{a}^{b} f(x) dx - \int_{a}^{b} g(x) dx \right|\)
Answer: False
In simple words: The area bounded by two curves \(y = f(x)\) and \(y = g(x)\) is \(\int_{a}^{b} |f(x) - g(x)| dx\), not necessarily involving the X-axis as a boundary for both functions directly in this form.

🎯 Exam Tip: The formula \(\int_{a}^{b} |f(x) - g(x)| dx\) calculates the area between two curves. The X-axis (y=0) is a specific case. The given expression is not always equivalent to the area bounded by two curves and the X-axis simultaneously, especially if the curves cross the X-axis or are entirely on one side.

Question 3. The area bounded by the curve y = f(x), X-axis and lines x = a and x = b is \(\int_{a}^{b} f(x) dx\)
Answer: True
In simple words: This is the fundamental definition of finding the area under a curve \(y = f(x)\) from \(x = a\) to \(x = b\) with respect to the X-axis.

🎯 Exam Tip: This is the most basic and frequently used formula for area calculation. Ensure the function \(f(x)\) is above the X-axis for the entire interval \([a, b]\) for the result to be directly the positive area.

Question 4. If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis, and lines x = a, x = b is positive.
Answer: False
In simple words: If a curve is below the X-axis, the definite integral will yield a negative value. Area, by definition, must be positive, so we take the absolute value of the integral.

🎯 Exam Tip: Area is always a positive quantity. If the integral results in a negative value, take its absolute value \(|\int_{a}^{b} f(x) dx|\) to represent the area.

Question 5. The area of the portion lying above the X-axis is positive.
Answer: True
In simple words: By convention and definition, regions above the X-axis (where \(y \ge 0\)) contribute positive values to the area calculated by integration.

🎯 Exam Tip: This reinforces the concept that area is a positive measure. If a function is positive, its integral represents the area above the axis; if negative, the integral represents the negative of the area below the axis.

(IV) Solve The Following:

Question 1. Find the area of the region bounded by the curve xy = c², the X-axis, and the lines x = c, x = 2c.
Solution:
Required area = \(\int_{c}^{2c} y dx\), where \(xy = c^2\), i.e. \(y = \frac{c^2}{x}\)
= \(\int_{c}^{2c} \frac{c^2}{x} dx\)
= \(c^2 \int_{c}^{2c} \frac{1}{x} dx\)
= \(c^2 [\log x]_{c}^{2c}\)
= \(c^2 (\log 2c - \log c)\)
= \(c^2 \log \left(\frac{2c}{c}\right)\)
= \(c^2 \log 2\) sq units.
In simple words: We found the area by integrating the function \(y = c^2/x\) from \(x = c\) to \(x = 2c\), resulting in \(c^2 \log 2\).

🎯 Exam Tip: Remember the integral of \(1/x\) is \(\log|x|\). Apply the limits carefully and use logarithm properties to simplify the final answer.

Question 2. Find the area between the parabolas y² = 7x and x² = 7y.
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो परवलयों \(y^2 = 7x\) और \(x^2 = 7y\) को दर्शाता है। परवलय X-अक्ष और Y-अक्ष को प्रतिच्छेद करते हैं, और O(0,0) और A(7,7) पर एक दूसरे को काटते हैं। आरेख इन दो वक्रों के बीच के क्षेत्र को छायांकित करके दिखाता है।
For finding the points of intersection of the two parabolas,
we equate the values of y² from their equations.
From the equation \(x^2 = 7y\), \(y = \frac{x^2}{7}\), so \(y^2 = \left(\frac{x^2}{7}\right)^2 = \frac{x^4}{49}\)
\(\frac{x^4}{49} = 7x\)

\(x^4 = 343x\)

\(x^4 - 343x = 0\)

\(x(x^3 - 343) = 0\)
\(\implies x = 0\) or \(x^3 = 343\), i.e. \(x = 7\)
When \(x = 0, y = 0\)
When \(x = 7, 7y = 49\)

\(y = 7\)
\(\implies\) the points of intersection are O(0, 0) and A(7, 7)
Required area = area of the region OBACO
= (area of the region ODACO) - (area of the region ODABO)
Now, area of the region ODACO = area under the parabola \(y^2 = 7x\)
i.e. \(y = \sqrt{7} \sqrt{x}\)
= \(\int_{0}^{7} \sqrt{7} \sqrt{x} dx = \sqrt{7} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{7}\)
= \(\sqrt{7} \times \frac{2}{3} [7^{3/2} - 0]\)
= \(\frac{2\sqrt{7}}{3} (7\sqrt{7} - 0)\)
= \(\frac{98}{3}\)
Area of the region ODABO = Area under the parabola \(x^2 = 7y\)
i.e. \(y = \frac{x^2}{7}\)
= \(\int_{0}^{7} \frac{x^2}{7} dx = \frac{1}{7} \left[ \frac{x^3}{3} \right]_{0}^{7}\)
= \(\frac{1}{7} \times \frac{1}{3} [7^3 - 0]\)
= \(\frac{7^2}{3} = \frac{49}{3}\)
\(\implies\) required area = \(\frac{98}{3} - \frac{49}{3} = \frac{49}{3}\) sq units.
In simple words: We found the intersection points of the two parabolas, then calculated the area under each curve separately and subtracted the smaller area from the larger one to get the area enclosed between them.

🎯 Exam Tip: For areas between curves, the general approach is \(\int_{a}^{b} (y_{upper} - y_{lower}) dx\) or \(\int_{c}^{d} (x_{right} - x_{left}) dy\). Correctly identifying the upper/lower or right/left curve is crucial.

Question 3. Find the area of the region bounded by the curve y = x² and the line y = 10.
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक ऊपर की ओर खुलने वाले परवलय \(x^2 = y\) और एक क्षैतिज रेखा \(y = 10\) को दर्शाता है। मूल O(0,0) है और रेखा परवलय को B और C बिंदुओं पर काटती है, जबकि A(0,10) Y-अक्ष पर है। इन वक्रों द्वारा घेरा गया क्षेत्र छायांकित है।
By the symmetry of the parabola,
the required area is twice the area of the region OABCO
Now, the area of the region OABCO
= \(\int_{0}^{10} x dy\), where \(x^2 = y\) i.e. \(x = \sqrt{y}\)
= \(\int_{0}^{10} \sqrt{y} dy\)
= \(\left[ \frac{y^{3/2}}{3/2} \right]_{0}^{10}\)
= \(\frac{2}{3} [10^{3/2} - 0]\)
= \(\frac{2}{3} \times 10\sqrt{10}\)
= \(\frac{20\sqrt{10}}{3}\)
\(\implies\) required area = \(2 \left[ \frac{20\sqrt{10}}{3} \right]\)
= \(\frac{40\sqrt{10}}{3}\) sq units.
In simple words: We found the area by integrating \(x = \sqrt{y}\) from \(y = 0\) to \(y = 10\) and then multiplied the result by 2 due to the symmetry of the parabola.

🎯 Exam Tip: Utilize symmetry whenever possible to simplify calculations. Integrating with respect to \(y\) can often be easier when the region is bounded by a horizontal line and a parabola opening along the X-axis.

Question 4. Find the area of the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख मूल पर केंद्रित एक दीर्घवृत्त को दर्शाता है जिसका समीकरण \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) है। दीर्घवृत्त X-अक्ष को P(4,0) पर और Y-अक्ष को Q(0,3) पर काटता है। प्रथम चतुर्थांश में दीर्घवृत्त द्वारा घेरा गया क्षेत्र (OPQO) छायांकित है, जिसका उपयोग कुल क्षेत्रफल की गणना के लिए किया जाता है।
By the symmetry of the ellipse, the required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.
Given equation of the ellipse is \(\frac{x^2}{16} + \frac{y^2}{9} = 1\)

\(\frac{y^2}{9} = 1 - \frac{x^2}{16}\)

\(y^2 = 9 \left(1 - \frac{x^2}{16}\right)\)

\(y^2 = \frac{9}{16} (16 - x^2)\)

\(y = \pm \frac{3}{4} \sqrt{16 - x^2}\)
\(\implies y = \frac{3}{4} \sqrt{16 - x^2}\) [: In first quadrant, \(y > 0\)]
\(\implies\) Required area = 4(area of the region OPQO)
= \(4 \int_{0}^{4} y dx\)
= \(4 \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} dx\)
= \(3 \int_{0}^{4} \sqrt{4^2 - x^2} dx\)
= \(3 \left[ \frac{x}{2} \sqrt{4^2 - x^2} + \frac{4^2}{2} \sin^{-1} \left(\frac{x}{4}\right) \right]_{0}^{4}\)
= \(3 \left\{ \left[ \frac{4}{2} \sqrt{4^2 - 4^2} + \frac{16}{2} \sin^{-1} \left(\frac{4}{4}\right) \right] - \left[ \frac{0}{2} \sqrt{4^2 - 0^2} + \frac{16}{2} \sin^{-1} \left(\frac{0}{4}\right) \right] \right\}\)
= \(3 \{ [0 + 8\sin^{-1} (1)] - [0 + 0] \}\)
= \(3 \left(8 \times \frac{\pi}{2}\right)\)
= \(12\pi\) sq. units.
In simple words: We used the symmetry of the ellipse to calculate the area of one quadrant and then multiplied by 4. The integration of \(\frac{3}{4}\sqrt{16-x^2}\) was performed using the standard formula for \(\int\sqrt{a^2-x^2} dx\).

🎯 Exam Tip: For standard ellipses, the area formula is \(\pi ab\). Here \(a=4\) and \(b=3\), so Area = \(\pi(4)(3) = 12\pi\). While direct integration shows the derivation, knowing the formula helps to verify your answer.

Question 5. Find the area of the region bounded by y = x², the X-axis and x = 1, x = 4.
Solution:
Required area = \(\int_{1}^{4} y dx\), where \(y = x^2\)
= \(\int_{1}^{4} x^2 dx\)
= \(\left[ \frac{x^3}{3} \right]_{1}^{4}\)
= \(\frac{1}{3} [4^3 - 1^3]\)
= \(\frac{1}{3} [64 - 1]\)
= \(\frac{63}{3}\)
= 21 sq units.
In simple words: The area is computed by performing a definite integral of the function \(y = x^2\) from the lower limit \(x=1\) to the upper limit \(x=4\).

🎯 Exam Tip: This is a straightforward area under the curve problem. Double-check the integral of \(x^2\) and the subtraction of the lower limit's value from the upper limit's value.

Question 6. Find the area of the region bounded by the curve x² = 25y, y = 1, y = 4, and the Y-axis.
Solution:
Required area = \(\int_{1}^{4} x dy\), where \(x^2 = 25y\), i.e. \(x = 5\sqrt{y}\)
= \(\int_{1}^{4} 5\sqrt{y} dy\)
= \(5 \int_{1}^{4} y^{1/2} dy\)
= \(5 \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4}\)
= \(\frac{10}{3} [4^{3/2} - 1^{3/2}]\)
= \(\frac{10}{3} [(\sqrt{4})^3 - (\sqrt{1})^3]\)
= \(\frac{10}{3} [2^3 - 1^3]\)
= \(\frac{10}{3} [8 - 1]\)
= \(\frac{10}{3} [7]\)
= \(\frac{70}{3}\) sq units.
In simple words: We calculated the area by integrating \(x = 5\sqrt{y}\) with respect to \(y\) from \(y=1\) to \(y=4\).

🎯 Exam Tip: When integrating with respect to \(y\), ensure \(x\) is expressed as a function of \(y\). Pay attention to the power rule for fractional exponents, e.g., \(y^{1/2}\).

Question 7. Find the area of the region bounded by the parabola y² = 25x and the line x = 5.
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक परवलय \(y^2 = 25x\) को दर्शाता है जो दाईं ओर खुलता है, और एक ऊर्ध्वाधर रेखा \(x = 5\) द्वारा प्रतिच्छेदित है। मूल O(0,0) है, और रेखा X-अक्ष को R(5,0) पर काटती है। परवलय और रेखा द्वारा घेरा गया क्षेत्र OQRPO के रूप में छायांकित है, जहाँ P और Q परवलय और रेखा के प्रतिच्छेद बिंदु हैं।
Given the equation of the parabola is \(y^2 = 25x\)
\(\implies y = \pm 5\sqrt{x}\)
\(\implies y = 5\sqrt{x}\) [... In first quadrant, \(y > 0\)]
Required area = area of the region OQRPO
= 2(area of the region ORPO)
= \(2 \int_{0}^{5} y dx\)
= \(2 \int_{0}^{5} 5\sqrt{x} dx\)
= \(10 \int_{0}^{5} x^{1/2} dx\)
= \(10 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{5}\)
= \(\frac{20}{3} [5^{3/2} - 0]\)
= \(\frac{20}{3} (5\sqrt{5})\)
= \(\frac{100\sqrt{5}}{3}\) sq.units.
In simple words: We used the symmetry of the parabola to calculate the area of the upper half by integrating \(y = 5\sqrt{x}\) from \(x=0\) to \(x=5\) and then multiplied by 2.

🎯 Exam Tip: Symmetry is a powerful tool to simplify area problems. For parabolas symmetric about the X-axis, calculate the area of the upper half and multiply by two.