Maharashtra Board Class 12 Maths Part 1 Chapter 5 Integration 5.1 Solutions

Question 1. Evaluate \( \int \frac{-2}{\sqrt{5x-4}-\sqrt{5x-2}} dx \)
Answer:
\( = \int \frac{-2}{\sqrt{5x-4} - \sqrt{5x-2}} \times \frac{\sqrt{5x-4} + \sqrt{5x-2}}{\sqrt{5x-4} + \sqrt{5x-2}} dx \)
\( = \int \frac{-2(\sqrt{5x-4} + \sqrt{5x-2})}{(5x-4)-(5x-2)} dx \)
\( = \int \frac{-2(\sqrt{5x-4} + \sqrt{5x-2})}{-2} dx \)
\( = \int (\sqrt{5x-4} + \sqrt{5x-2}) dx \)
\( = \int (5x-4)^{1/2} dx + \int (5x-2)^{1/2} dx \)
\( = \frac{(5x-4)^{3/2}}{\frac{3}{2} \times 5} + \frac{(5x-2)^{3/2}}{\frac{3}{2} \times 5} + c \)
\( = \frac{2}{15}(5x-4)^{3/2} + \frac{2}{15}(5x-2)^{3/2} + c \)
\( = \frac{2}{15} [(5x-4)^{3/2} + (5x-2)^{3/2}] + c \) In simple words: This problem is solved by rationalizing the denominator, then applying the power rule for integration for each term.

🎯 Exam Tip: Remember to rationalize the denominator when integrals involve square roots in the denominator, and always apply the chain rule adjustment for the derivative of the inner function (e.g., \( \frac{1}{5} \) for \( (5x-4) \)).

Question 2. Evaluate \( \int (1 + x + \frac{x^2}{2!}) dx \)
Answer:
\( = \int 1 dx + \int x dx + \int \frac{x^2}{2} dx \)
\( = x + \frac{x^2}{2} + \frac{1}{2} \times \frac{x^3}{3} + c \)
\( = x + \frac{x^2}{2} + \frac{x^3}{6} + c \) In simple words: This integral is solved by integrating each term separately using the basic power rule of integration.

🎯 Exam Tip: For polynomial functions, integrate each term individually. Don't forget the constant of integration, 'c', for indefinite integrals.

Question 3. Evaluate \( \int \frac{3x^3-2\sqrt{x}}{x} dx \)
Answer:
\( = \int (\frac{3x^3}{x} - \frac{2\sqrt{x}}{x}) dx \)
\( = \int (3x^2 - \frac{2x^{1/2}}{x}) dx \)
\( = \int (3x^2 - 2x^{-1/2}) dx \)
\( = 3 \int x^2 dx - 2 \int x^{-1/2} dx \)
\( = 3 \times \frac{x^3}{3} - 2 \times \frac{x^{1/2}}{1/2} + c \)
\( = x^3 - 4x^{1/2} + c \)
\( = x^3 - 4\sqrt{x} + c \) In simple words: Simplify the integrand by dividing each term in the numerator by \( x \), then integrate using the power rule.

🎯 Exam Tip: Always simplify algebraic expressions before integrating. Express square roots as fractional exponents to easily apply the power rule for integration.

Question 4. Evaluate \( \int (3x^2 - 5)^2 dx \)
Answer:
\( = \int (9x^4 - 30x^2 + 25) dx \)
\( = 9 \int x^4 dx - 30 \int x^2 dx + 25 \int 1 dx \)
\( = 9 \times \frac{x^5}{5} - 30 \times \frac{x^3}{3} + 25x + c \)
\( = \frac{9x^5}{5} - 10x^3 + 25x + c \) In simple words: Expand the squared term, then integrate each resulting polynomial term separately using the power rule.

🎯 Exam Tip: When integrating squared binomials, first expand the expression. Then, apply the linearity of integration and the power rule to each term.

Question 5. Evaluate \( \int \frac{1}{x(x-1)} dx \)
Answer:
\( = \int (\frac{1}{x-1} - \frac{1}{x}) dx \)
\( = \int \frac{1}{x-1} dx - \int \frac{1}{x} dx \)
\( = \log |x-1| - \log |x| + c \)
\( = \log |\frac{x-1}{x}| + c \) In simple words: Decompose the fraction into simpler partial fractions, then integrate each resulting logarithmic term.

🎯 Exam Tip: For rational functions, consider using partial fraction decomposition to simplify the integrand into terms that are easier to integrate, typically leading to logarithmic functions.

Question 6. If \( f'(x) = x^2 + 5 \) and \( f(0) = -1 \), then find the value of \( f(x) \).
Answer: By the definition of integral \( f(x) = \int f'(x) dx \)
\( = \int (x^2 + 5) dx \)
\( = \int x^2 dx + 5 \int 1 dx \)
\( = \frac{x^3}{3} + 5x + c \) Now, \( f(0) = -1 \) gives \( f(0) = 0 + 0 + c = -1 \)
\( \implies c = -1 \) Therefore, \( f(x) = \frac{x^3}{3} + 5x - 1 \). In simple words: Integrate the given derivative to find the function, then use the initial condition \( f(0) = -1 \) to determine the constant of integration.

🎯 Exam Tip: Remember to always solve for the constant of integration 'c' using the given initial condition. A common mistake is to forget this crucial step.

Question 7. If \( f'(x) = 4x^3 - 3x^2 + 2x + k \), \( f(0) = -1 \) and \( f(1) = 4 \), find \( f(x) \).
Answer: By the definition of integral \( f(x) = \int f'(x) dx \)
\( = \int (4x^3 - 3x^2 + 2x + k) dx \)
\( = 4 \int x^3 dx - 3 \int x^2 dx + 2 \int x dx + k \int 1 dx \)
\( = 4 \times \frac{x^4}{4} - 3 \times \frac{x^3}{3} + 2 \times \frac{x^2}{2} + kx + c \) Therefore, \( f(x) = x^4 - x^3 + x^2 + kx + c \) Now, \( f(0) = 1 \) gives \( f(0) = 0 - 0 + 0 + 0 + c = 1 \)
\( \implies c = 1 \) Therefore, \( f(x) = x^4 - x^3 + x^2 + kx + 1 \) Further \( f(1) = 4 \) gives \( f(1) = 1 - 1 + 1 + k + 1 = 4 \)
\( \implies k = 2 \) Therefore, \( f(x) = x^4 - x^3 + x^2 + 2x + 1 \). In simple words: Integrate the given derivative to find the general function \( f(x) \), then use the two given conditions, \( f(0) = 1 \) and \( f(1) = 4 \), to solve for both constants \( c \) and \( k \).

🎯 Exam Tip: For problems with multiple unknown constants (like 'k' and 'c'), you'll need an equal number of initial conditions to solve for them. Substitute the values carefully to avoid errors.

Question 8. If \( f'(x) = \frac{x^2}{2} - kx + 1 \), \( f(0) = 2 \) and \( f(3) = 5 \), find \( f(x) \).
Answer: By the definition of integral \( f(x) = \int f'(x) dx \)
\( = \int (\frac{x^2}{2} - kx + 1) dx \)
\( = \frac{1}{2} \int x^2 dx - k \int x dx + \int 1 dx \)
\( = \frac{1}{2} \times \frac{x^3}{3} - k \times \frac{x^2}{2} + x + c \) Therefore, \( f(x) = \frac{x^3}{6} - \frac{kx^2}{2} + x + c \) Now, \( f(0) = 2 \) gives \( f(0) = 0 - 0 + 0 + c = 2 \)
\( \implies c = 2 \) Therefore, \( f(x) = \frac{x^3}{6} - \frac{kx^2}{2} + x + 2 \) Further \( f(3) = 5 \) gives \( f(3) = \frac{3^3}{6} - \frac{k(3^2)}{2} + 3 + 2 = 5 \) \( \frac{27}{6} - \frac{9k}{2} + 5 = 5 \) \( \frac{9}{2} - \frac{9k}{2} = 0 \)
\( \implies \frac{9}{2} (1-k) = 0 \)
\( \implies 1-k=0 \)
\( \implies k = 1 \) Therefore, \( f(x) = \frac{x^3}{6} - \frac{x^2}{2} + x + 2 \). In simple words: Integrate the derivative to find the general function \( f(x) \), then use the first condition \( f(0) = 2 \) to find \( c \), and finally use \( f(3) = 5 \) to find the value of \( k \).

🎯 Exam Tip: When given multiple conditions, use them sequentially. First, find the integration constant, then use the second condition to solve for any other unknown coefficients in the function.