Maharashtra Board Class 12 Maths Part 1 Chapter 7 Application of Definite 7.1 Solutions

Question 1. Find the area of the region bounded by the following curves, the X-axis, and the given lines:
(i) y = x⁴, x = 1, x = 5
Answer: Solution:
Required area = \( \int_1^5 y \, dx \), where \( y = x^4 \)
= \( \int_1^5 x^4 \, dx \)
= \( \left[ \frac{x^5}{5} \right]_1^5 \)
= \( \frac{1}{5} [5^5 - 1^5] \)
= \( \frac{1}{5} [3125 - 1] \)
= \( \frac{3124}{5} \) sq units.
In simple words: To find the area, we integrate the function y=x⁴ with respect to x from the lower limit x=1 to the upper limit x=5, then substitute the limits into the antiderivative and calculate the difference.

🎯 Exam Tip: Remember to apply the correct integration limits and correctly evaluate the definite integral for scoring maximum marks.

(ii) y = \( \sqrt{6x + 4} \), x = 0, x = 2
Answer: Solution:
Required area = \( \int_0^2 y \, dx \), where \( y = \sqrt{6x+4} \)
= \( \int_0^2 \sqrt{6x+4} \, dx \)
= \( \int_0^2 (6x+4)^{1/2} \, dx \)
= \( \left[ \frac{(6x+4)^{3/2}}{3/2} \cdot \frac{1}{6} \right]_0^2 \)
= \( \left[ \frac{(6x+4)^{3/2}}{9} \right]_0^2 \)
= \( \frac{1}{9} [(6(2)+4)^{3/2} - (6(0)+4)^{3/2}] \)
= \( \frac{1}{9} [(12+4)^{3/2} - (4)^{3/2}] \)
= \( \frac{1}{9} [16^{3/2} - 4^{3/2}] \)
= \( \frac{1}{9} [(\sqrt{16})^3 - (\sqrt{4})^3] \)
= \( \frac{1}{9} [4^3 - 2^3] \)
= \( \frac{1}{9} [64 - 8] \)
= \( \frac{56}{9} \) sq units.
In simple words: The area under the curve is found by integrating \( \sqrt{6x+4} \) from x=0 to x=2, using the power rule for integration and evaluating at the given limits.

🎯 Exam Tip: Pay close attention to the exponent when integrating square roots and ensure correct substitution of limits to avoid calculation errors.

(iii) y = \( \sqrt{16 - x^2} \), x = 0, x = 4
Answer: Solution:
Required area = \( \int_0^4 y \, dx \), where \( y = \sqrt{16-x^2} \)
= \( \int_0^4 \sqrt{4^2-x^2} \, dx \)
We use the formula: \( \int \sqrt{a^2-x^2} \, dx = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) \)
Here, \( a = 4 \).
= \( \left[ \frac{x}{2} \sqrt{16-x^2} + \frac{16}{2} \sin^{-1} \left( \frac{x}{4} \right) \right]_0^4 \)
= \( \left[ \frac{4}{2} \sqrt{16-4^2} + 8 \sin^{-1} \left( \frac{4}{4} \right) \right] - \left[ \frac{0}{2} \sqrt{16-0^2} + 8 \sin^{-1} \left( \frac{0}{4} \right) \right] \)
= \( [2 \sqrt{16-16} + 8 \sin^{-1} (1)] - [0 + 8 \sin^{-1} (0)] \)
= \( [2 \cdot 0 + 8 \cdot \frac{\pi}{2}] - [0 + 8 \cdot 0] \)
= \( 0 + 8 \cdot \frac{\pi}{2} - 0 \)
= \( 4\pi \) sq units.
In simple words: This problem involves integrating a semicircle's equation. We use the standard formula for \( \int \sqrt{a^2-x^2} \, dx \), substituting the limits to find the area under the curve.

🎯 Exam Tip: Recognizing the form \( \sqrt{a^2-x^2} \) is key, as it indicates the use of a specific integration formula, which must be memorized for accurate solutions.

(iv) 2y = 5x + 7, x = 2, x = 8
Answer: Solution:
From the given equation, \( 2y = 5x+7 \), so \( y = \frac{5x+7}{2} \).
Required area = \( \int_2^8 y \, dx \)
= \( \int_2^8 \frac{5x+7}{2} \, dx \)
= \( \frac{1}{2} \int_2^8 (5x+7) \, dx \)
= \( \frac{1}{2} \left[ \frac{5x^2}{2} + 7x \right]_2^8 \)
= \( \frac{1}{2} \left[ \left( \frac{5(8)^2}{2} + 7(8) \right) - \left( \frac{5(2)^2}{2} + 7(2) \right) \right] \)
= \( \frac{1}{2} \left[ \left( \frac{5 \cdot 64}{2} + 56 \right) - \left( \frac{5 \cdot 4}{2} + 14 \right) \right] \)
= \( \frac{1}{2} \left[ (5 \cdot 32 + 56) - (5 \cdot 2 + 14) \right] \)
= \( \frac{1}{2} \left[ (160 + 56) - (10 + 14) \right] \)
= \( \frac{1}{2} [216 - 24] \)
= \( \frac{1}{2} [192] \)
= 96 sq units.
In simple words: We find the area under the line \( y = \frac{5x+7}{2} \) by integrating it from x=2 to x=8, applying the power rule for integration, and then subtracting the lower limit evaluation from the upper limit evaluation.

🎯 Exam Tip: Simplify the equation for y first, then perform the integration carefully, paying attention to the coefficients and limits to minimize errors.

(v) 2y + x = 8, x = 2, x = 4
Answer: Solution:
From the given equation, \( 2y+x=8 \), so \( 2y = 8-x \), and \( y = \frac{8-x}{2} \).
Required area = \( \int_2^4 y \, dx \)
= \( \int_2^4 \frac{8-x}{2} \, dx \)
= \( \frac{1}{2} \int_2^4 (8-x) \, dx \)
= \( \frac{1}{2} \left[ 8x - \frac{x^2}{2} \right]_2^4 \)
= \( \frac{1}{2} \left[ \left( 8(4) - \frac{4^2}{2} \right) - \left( 8(2) - \frac{2^2}{2} \right) \right] \)
= \( \frac{1}{2} \left[ \left( 32 - \frac{16}{2} \right) - \left( 16 - \frac{4}{2} \right) \right] \)
= \( \frac{1}{2} \left[ (32 - 8) - (16 - 2) \right] \)
= \( \frac{1}{2} [24 - 14] \)
= \( \frac{1}{2} [10] \)
= 5 sq units.
In simple words: The area is calculated by integrating the linear function \( y = \frac{8-x}{2} \) with respect to x, from x=2 to x=4, then applying the fundamental theorem of calculus.

🎯 Exam Tip: Always rearrange the equation to express y explicitly in terms of x before integrating to simplify the process and reduce errors.

(vi) y = x² + 1, x = 0, x = 3
Answer: Solution:
Required area = \( \int_0^3 y \, dx \), where \( y = x^2+1 \)
= \( \int_0^3 (x^2+1) \, dx \)
= \( \left[ \frac{x^3}{3} + x \right]_0^3 \)
= \( \left( \frac{3^3}{3} + 3 \right) - \left( \frac{0^3}{3} + 0 \right) \)
= \( \left( \frac{27}{3} + 3 \right) - (0 + 0) \)
= \( (9 + 3) - 0 \)
= 12 sq units.
In simple words: To find the area, we integrate the parabolic function \( y = x^2+1 \) from x=0 to x=3, using the power rule for integration, and then evaluate the definite integral.

🎯 Exam Tip: Standard polynomial integration is straightforward; ensure careful calculation of powers and constants, especially when evaluating at the limits.

(vii) y = 2 - x², x = -1, x = 1
Answer: Solution:
Required area = \( \int_{-1}^1 y \, dx \), where \( y = 2-x^2 \)
= \( \int_{-1}^1 (2-x^2) \, dx \)
Since \( f(x) = 2-x^2 \) is an even function (i.e., \( f(-x) = f(x) \)), we can write:
= \( 2 \int_0^1 (2-x^2) \, dx \)
= \( 2 \left[ 2x - \frac{x^3}{3} \right]_0^1 \)
= \( 2 \left[ \left( 2(1) - \frac{1^3}{3} \right) - \left( 2(0) - \frac{0^3}{3} \right) \right] \)
= \( 2 \left[ \left( 2 - \frac{1}{3} \right) - (0 - 0) \right] \)
= \( 2 \left[ \frac{6-1}{3} \right] \)
= \( 2 \left[ \frac{5}{3} \right] \)
= \( \frac{10}{3} \) sq units.
In simple words: The area is found by integrating the function \( y = 2-x^2 \) from x=-1 to x=1. Since the function is even, we can integrate from 0 to 1 and multiply the result by 2 for simplicity.

🎯 Exam Tip: Identifying even or odd functions can simplify calculations by allowing you to change integration limits, especially over symmetric intervals like \([-a, a]\).

Question 2. Find the area of the region bounded by the parabola y² = 4x and the line x = 3.
Answer: Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र x-अक्ष पर सममित एक परवलय \( y^2 = 4x \) और x=3 पर एक ऊर्ध्वाधर रेखा को दर्शाता है। यह परवलय मूल बिंदु (0,0) से शुरू होकर धनात्मक x-अक्ष की ओर खुलता है। रेखा x=3, x-अक्ष को C(3,0) पर काटती है और परवलय को बिंदु A और B पर काटती है। छायांकित क्षेत्र OABO रेखा x=3 और परवलय द्वारा परिबद्ध क्षेत्र है, जिसका क्षेत्रफल ज्ञात करना है। Required area = area of the region OABO
= 2(area of the region OACO)
From the equation of the parabola, \( y^2 = 4x \), so \( y = \sqrt{4x} = 2\sqrt{x} \).
Limits of integration are from x = 0 to x = 3.
Area of region OACO = \( \int_0^3 y \, dx \)
= \( \int_0^3 2\sqrt{x} \, dx \)
= \( 2 \int_0^3 x^{1/2} \, dx \)
= \( 2 \left[ \frac{x^{1/2+1}}{1/2+1} \right]_0^3 \)
= \( 2 \left[ \frac{x^{3/2}}{3/2} \right]_0^3 \)
= \( 2 \cdot \frac{2}{3} [x^{3/2}]_0^3 \)
= \( \frac{4}{3} [3^{3/2} - 0^{3/2}] \)
= \( \frac{4}{3} [3\sqrt{3} - 0] \)
= \( \frac{4}{3} \cdot 3\sqrt{3} \)
= \( 4\sqrt{3} \) sq units.
Required area = 2 * (Area of OACO)
= \( 2 \cdot 4\sqrt{3} \)
= \( 8\sqrt{3} \) sq units.
In simple words: The area bounded by the parabola \( y^2=4x \) and the line \( x=3 \) is symmetric about the x-axis. We find the area in the first quadrant by integrating \( y = 2\sqrt{x} \) from 0 to 3, and then double it to get the total area.

🎯 Exam Tip: For symmetric regions, calculating the area in one quadrant and multiplying by the number of symmetric parts can save significant calculation time and reduce errors.

Question 3. Find the area of the circle x² + y² = 25.
Answer: Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र केंद्र (0,0) पर और 5 इकाई त्रिज्या वाले एक वृत्त को दर्शाता है, जिसका समीकरण \( x^2 + y^2 = 25 \) है। यह वृत्त x-अक्ष को (5,0) पर A और (-5,0) पर, और y-अक्ष को (0,5) पर B और (0,-5) पर काटता है। छायांकित क्षेत्र पहले चतुर्थांश में वृत्त का वह हिस्सा है जो x=0 से x=5 तक फैला हुआ है। वृत्त की समरूपता के कारण, इस पूरे वृत्त का क्षेत्रफल पहले चतुर्थांश में इस छायांकित क्षेत्र के क्षेत्रफल का चार गुना होगा। By the symmetry of the circle, its area is equal to 4 times the area of the region OABO (area in the first quadrant).
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the circle, \( x^2+y^2=25 \), so \( y^2 = 25-x^2 \).
In the first quadrant, \( y > 0 \), so \( y = \sqrt{25-x^2} \).
Area of region OABO = \( \int_0^5 y \, dx \)
= \( \int_0^5 \sqrt{25-x^2} \, dx \)
We use the formula: \( \int \sqrt{a^2-x^2} \, dx = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) \)
Here, \( a = 5 \).
= \( \left[ \frac{x}{2} \sqrt{25-x^2} + \frac{25}{2} \sin^{-1} \left( \frac{x}{5} \right) \right]_0^5 \)
= \( \left[ \frac{5}{2} \sqrt{25-5^2} + \frac{25}{2} \sin^{-1} \left( \frac{5}{5} \right) \right] - \left[ \frac{0}{2} \sqrt{25-0^2} + \frac{25}{2} \sin^{-1} \left( \frac{0}{5} \right) \right] \)
= \( \left[ \frac{5}{2} \cdot 0 + \frac{25}{2} \sin^{-1} (1) \right] - \left[ 0 + \frac{25}{2} \sin^{-1} (0) \right] \)
= \( \left[ 0 + \frac{25}{2} \cdot \frac{\pi}{2} \right] - [0 + 0] \)
= \( \frac{25\pi}{4} \) sq units.
Therefore, the total area of the circle = 4 * (Area of OABO)
= \( 4 \cdot \frac{25\pi}{4} \)
= \( 25\pi \) sq units.
In simple words: The area of the circle is found by calculating the area in the first quadrant using integration of \( y = \sqrt{25-x^2} \) from x=0 to x=5, and then multiplying this result by 4 due to the circle's symmetry.

🎯 Exam Tip: Utilizing symmetry for closed figures like circles or ellipses can significantly reduce the integration limits and complexity of calculations. Remember the formula for the area of a circle, \( \pi r^2 \), and ensure your calculated r matches the given equation.

Question 4. Find the area of the ellipse \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \).
Answer: Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक दीर्घवृत्त को दर्शाता है जिसका केंद्र मूल बिंदु (0,0) पर है। इसका समीकरण \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \) है। यह दीर्घवृत्त x-अक्ष को (2,0) पर A और (-2,0) पर, और y-अक्ष को (0,5) पर B और (0,-5) पर काटता है। छायांकित क्षेत्र पहले चतुर्थांश में दीर्घवृत्त का वह हिस्सा है जो x=0 से x=2 तक फैला हुआ है। दीर्घवृत्त की समरूपता के कारण, इस पूरे दीर्घवृत्त का क्षेत्रफल पहले चतुर्थांश में इस छायांकित क्षेत्र के क्षेत्रफल का चार गुना होगा। By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO (area in the first quadrant).
Clearly, for this region, the limits of integration are 0 and 2.
From the equation of the ellipse, \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \).
We need to express \( y \) in terms of \( x \):
\( \frac{y^2}{25} = 1 - \frac{x^2}{4} \)
\( y^2 = 25 \left( 1 - \frac{x^2}{4} \right) \)
\( y^2 = \frac{25}{4} (4 - x^2) \)
In the first quadrant, \( y > 0 \), so \( y = \sqrt{\frac{25}{4} (4 - x^2)} = \frac{5}{2} \sqrt{4 - x^2} \).
Area of region OABO = \( \int_0^2 y \, dx \)
= \( \int_0^2 \frac{5}{2} \sqrt{4-x^2} \, dx \)
= \( \frac{5}{2} \int_0^2 \sqrt{2^2-x^2} \, dx \)
We use the formula: \( \int \sqrt{a^2-x^2} \, dx = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) \)
Here, \( a = 2 \).
= \( \frac{5}{2} \left[ \frac{x}{2} \sqrt{4-x^2} + \frac{4}{2} \sin^{-1} \left( \frac{x}{2} \right) \right]_0^2 \)
= \( \frac{5}{2} \left[ \left( \frac{2}{2} \sqrt{4-2^2} + 2 \sin^{-1} \left( \frac{2}{2} \right) \right) - \left( \frac{0}{2} \sqrt{4-0^2} + 2 \sin^{-1} \left( \frac{0}{2} \right) \right) \right] \)
= \( \frac{5}{2} \left[ \left( 1 \cdot 0 + 2 \sin^{-1} (1) \right) - \left( 0 + 2 \sin^{-1} (0) \right) \right] \)
= \( \frac{5}{2} \left[ \left( 0 + 2 \cdot \frac{\pi}{2} \right) - (0 + 2 \cdot 0) \right] \)
= \( \frac{5}{2} \left[ \pi - 0 \right] \)
= \( \frac{5\pi}{2} \) sq units.
Therefore, the total area of the ellipse = 4 * (Area of OABO)
= \( 4 \cdot \frac{5\pi}{2} \)
= \( 10\pi \) sq units.
In simple words: The area of the ellipse is found by integrating the upper half of the ellipse in the first quadrant from x=0 to x=2, then multiplying the result by 4 due to the ellipse's four-fold symmetry.

🎯 Exam Tip: Remember to solve the ellipse equation for y and correctly identify 'a' for the integration formula. The total area of an ellipse is \( \pi ab \), which can be used as a quick check for your calculated answer.