Maharashtra Board Class 12 Maths Part 1 Chapter 6 Definite Integration 6.2 Solutions

Evaluate the following integrals:

Question 1. \( \int_{-9}^{9} \frac{x^3}{4-x^2} dx \)
Answer: Solution: Let \( I = \int_{-9}^{9} \frac{x^3}{4-x^2} dx \) Let \( f(x) = \frac{x^3}{4-x^2} \) \( \implies f(-x) = \frac{(-x)^3}{4-(-x)^2} = \frac{-x^3}{4+x^2} = -f(x) \) \( \implies f \) is an odd function. Therefore, \( \int_{-9}^{9} f(x) dx = 0 \) i.e., \( \int_{-9}^{9} \frac{x^3}{4-x^2} dx = 0 \).
In simple words: This problem utilizes the property of definite integrals for odd functions over a symmetric interval \([-a, a]\), where the integral of an odd function is always zero.

🎯 Exam Tip: Always check the integrand's parity (odd or even) when the integration limits are symmetric about zero (e.g., from \(-a\) to \(a\)).

Question 2. \( \int_{0}^{a} x^2 (a - x)^{3/2} dx \)
Answer: Solution: We use the property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \) Let \( I = \int_{0}^{a} x^2 (a-x)^{3/2} dx \)
\( \implies I = \int_{0}^{a} (a-x)^2 (a-(a-x))^{3/2} dx \)
\( = \int_{0}^{a} (a-x)^2 x^{3/2} dx \)
\( = \int_{0}^{a} (a^2 - 2ax + x^2) x^{3/2} dx \)
\( = \int_{0}^{a} (a^2 x^{3/2} - 2ax^{5/2} + x^{7/2}) dx \)
\( = a^2 \int_{0}^{a} x^{3/2} dx - 2a \int_{0}^{a} x^{5/2} dx + \int_{0}^{a} x^{7/2} dx \)
\( = a^2 \left[ \frac{x^{5/2}}{5/2} \right]_{0}^{a} - 2a \left[ \frac{x^{7/2}}{7/2} \right]_{0}^{a} + \left[ \frac{x^{9/2}}{9/2} \right]_{0}^{a} \)
\( = a^2 \left( \frac{2}{5} a^{5/2} - 0 \right) - 2a \left( \frac{2}{7} a^{7/2} - 0 \right) + \left( \frac{2}{9} a^{9/2} - 0 \right) \)
\( = \frac{2}{5} a^{9/2} - \frac{4}{7} a^{9/2} + \frac{2}{9} a^{9/2} \)
\( = a^{9/2} \left( \frac{2}{5} - \frac{4}{7} + \frac{2}{9} \right) \)
\( = a^{9/2} \left( \frac{2 \times 63 - 4 \times 45 + 2 \times 35}{315} \right) \)
\( = a^{9/2} \left( \frac{126 - 180 + 70}{315} \right) \)
\( = \frac{16}{315} a^{9/2} \).
In simple words: This problem demonstrates how to use the property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \) to simplify the integrand into a polynomial form, making the integration straightforward.

🎯 Exam Tip: Remember to apply the property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \) when you see integrals with terms like \( x^m (a-x)^n \), as it often simplifies the expression for integration.

Question 3. \( \int_{1}^{3} \frac{\sqrt{x+5}}{\sqrt{x+5}+\sqrt{9-x}} dx \)
Answer: Solution: Let \( I = \int_{1}^{3} \frac{\sqrt{x+5}}{\sqrt{x+5}+\sqrt{9-x}} dx \) ... (1) We use the property, \( \int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx \) Hence in I, we replace \( x \) by \( 1+3-x = 4-x \).
\( \implies I = \int_{1}^{3} \frac{\sqrt{(4-x)+5}}{\sqrt{(4-x)+5}+\sqrt{9-(4-x)}} dx \)
\( = \int_{1}^{3} \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{5+x}} dx \) ... (2) Adding (1) and (2), we get
\( 2I = \int_{1}^{3} \left( \frac{\sqrt{x+5}}{\sqrt{x+5}+\sqrt{9-x}} + \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x+5}} \right) dx \)
\( = \int_{1}^{3} \frac{\sqrt{x+5}+\sqrt{9-x}}{\sqrt{x+5}+\sqrt{9-x}} dx \)
\( = \int_{1}^{3} 1 dx = [x]_{1}^{3} \)
\( = 3-1 = 2 \)
\( \implies I = 1 \) Hence, \( \int_{1}^{3} \frac{\sqrt{x+5}}{\sqrt{x+5}+\sqrt{9-x}} dx = 1 \).
In simple words: This problem is solved by applying a standard definite integral property where replacing \(x\) with \((a+b-x)\) makes the numerator and denominator identical, simplifying the integrand to 1.

🎯 Exam Tip: For integrals of the form \( \int_{a}^{b} \frac{f(x)}{f(x)+f(a+b-x)} dx \), the result is always \( \frac{b-a}{2} \). In this case, \( f(x) = \sqrt{x+5} \) and \( f(a+b-x) = f(4-x) = \sqrt{9-x} \), so the integral simplifies to \( \frac{3-1}{2} = 1 \).

Question 4. \( \int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} dx \)
Answer: Solution: Let \( I = \int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} dx \) ... (1) We use the property, \( \int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx \) Hence in I, we change \( x \) by \( 2+5-x = 7-x \).
\( \implies I = \int_{2}^{5} \frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{7-(7-x)}} dx \)
\( = \int_{2}^{5} \frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}} dx \) ... (2) Adding (1) and (2), we get
\( 2I = \int_{2}^{5} \left( \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} + \frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}} \right) dx \)
\( = \int_{2}^{5} \frac{\sqrt{x}+\sqrt{7-x}}{\sqrt{x}+\sqrt{7-x}} dx \)
\( = \int_{2}^{5} 1 dx = [x]_{2}^{5} \)
\( = 5-2 = 3 \)
\( \implies I = \frac{3}{2} \) Hence, \( \int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}} dx = \frac{3}{2} \).
In simple words: By applying the property \( \int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx \), the integral simplifies to a form where the numerator and denominator are identical, allowing for easy evaluation.

🎯 Exam Tip: This type of integral commonly appears in exams. Recognize the structure \( \int_{a}^{b} \frac{g(x)}{g(x)+g(a+b-x)} dx \) where the result is always \( \frac{b-a}{2} \). Here, \( \frac{5-2}{2} = \frac{3}{2} \).

Question 5. \( \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} dx \)
Answer: Solution: Let \( I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} dx \) ... (i)
\( \implies \) We use the property \( \int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx \)
\( \implies \) Hence in I, we replace \( x \) by \( 1+2-x = 3-x \).
\( \implies I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{3-(3-x)}+\sqrt{3-x}} dx \)
\( = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} dx \) ... (ii) Adding (i) and (ii), we get
\( 2I = \int_{1}^{2} \left( \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} + \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} \right) dx \)
\( = \int_{1}^{2} \frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} dx \)
\( = \int_{1}^{2} 1 \cdot dx = [x]_{1}^{2} \)
\( \implies 2I = 2-1 = 1 \)
\( \implies I = \frac{1}{2} \).
In simple words: This problem is solved using the property \( \int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx \). By replacing \(x\) with \((a+b-x)\), the integrand transforms into a form that, when added to the original, simplifies to 1.

🎯 Exam Tip: Integrals of the type \( \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{A-x}+\sqrt{x}} dx \) where \(A=a+b\) can often be solved quickly by remembering that the result is \( \frac{b-a}{2} \). Here, \( \frac{2-1}{2} = \frac{1}{2} \).

Question 6. \( \int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} dx \)
Answer: Solution: Let \( I = \int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} dx \) ... (i) We use the property, \( \int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx \) Hence in I, we replace \( x \) by \( 2+7-x = 9-x \).
\( \implies I = \int_{2}^{7} \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{9-(9-x)}} dx \)
\( = \int_{2}^{7} \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} dx \) ... (ii) Adding (i) and (ii), we get
\( 2I = \int_{2}^{7} \left( \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} + \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} \right) dx \)
\( = \int_{2}^{7} \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} dx \)
\( = \int_{2}^{7} 1 \cdot dx = [x]_{2}^{7} \)
\( \implies 2I = 7-2 = 5 \)
\( \implies I = \frac{5}{2} \).
In simple words: This integral is efficiently solved by applying the property \( \int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx \), which transforms the integrand into a form that simplifies to 1 when added to the original integral.

🎯 Exam Tip: This is a common pattern for definite integrals. If the integrand is of the form \( \frac{g(x)}{g(x)+g(a+b-x)} \), the value of the integral is simply \( \frac{b-a}{2} \). Here, \( \frac{7-2}{2} = \frac{5}{2} \).

Question 7. \( \int_{0}^{1} \log \left( \frac{x}{1-x} - 1 \right) dx \)
Answer: Solution: Let \( I = \int_{0}^{1} \log \left( \frac{1}{x} - 1 \right) dx = \int_{0}^{1} \log \left( \frac{1-x}{x} \right) dx \) We use the property, \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \)
\( \implies I = \int_{0}^{1} \log \left( \frac{1-(1-x)}{1-x} \right) dx \)
\( = \int_{0}^{1} \log \left( \frac{x}{1-x} \right) dx \) Adding the two forms of I:
\( 2I = \int_{0}^{1} \left[ \log \left( \frac{1-x}{x} \right) + \log \left( \frac{x}{1-x} \right) \right] dx \)
\( 2I = \int_{0}^{1} \log \left( \frac{1-x}{x} \cdot \frac{x}{1-x} \right) dx \)
\( 2I = \int_{0}^{1} \log(1) dx \)
\( 2I = \int_{0}^{1} 0 dx = 0 \)
\( \implies I = 0 \) Hence, \( \int_{0}^{1} \log \left( \frac{1}{x} - 1 \right) dx = 0 \).
In simple words: This integral is solved by using the property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \). By adding the original and transformed integrals, the logarithmic terms combine to form \( \log(1) \), which is zero, simplifying the entire integral.

🎯 Exam Tip: When integrating logarithmic functions, especially with limits from 0 to 'a', always consider the property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \) as it often leads to significant simplifications through the properties of logarithms.

Question 8. \( \int_{0}^{1} x(1-x)^5 dx \)
Answer: Solution: We use the property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \). Let \( I = \int_{0}^{1} x(1-x)^5 dx \)
\( \implies I = \int_{0}^{1} (1-x)(1-(1-x))^5 dx \)
\( = \int_{0}^{1} (1-x)x^5 dx \)
\( = \int_{0}^{1} (x^5 - x^6) dx \)
\( = \left[ \frac{x^6}{6} - \frac{x^7}{7} \right]_{0}^{1} \)
\( = \left( \frac{1^6}{6} - \frac{1^7}{7} \right) - (0 - 0) \)
\( = \frac{1}{6} - \frac{1}{7} \)
\( = \frac{7-6}{42} = \frac{1}{42} \).
In simple words: This problem efficiently uses the property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \) to transform the integral of a product into the integral of a simple polynomial, which is then easy to solve.

🎯 Exam Tip: For integrals of the form \( \int_{0}^{a} x^m (a-x)^n dx \), applying the property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \) simplifies the integrand, often reducing it to a straightforward polynomial integration.