Maharashtra Board Class 12 Maths Part 1 Chapter 6 Definite Integration 6.1 Solutions

Evaluate The Following Definite Integrals:

Question 1. \( \int_{1}^{9} \frac{1}{\sqrt{x}} dx \)
Answer: Solution: \( \int_{4}^{9} \frac{1}{\sqrt{x}} dx = \int_{4}^{9} x^{- \frac{1}{2}} dx \) \( = \left[ \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right]_{4}^{9} = \left[ \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{4}^{9} = 2 [ \sqrt{x} ]_{4}^{9} \) \( = 2(\sqrt{9} - \sqrt{4}) \) \( = 2(3 - 2) = 2. \) In simple words: To evaluate this definite integral, we first express the square root in the denominator as a fractional exponent, integrate the power function, and then apply the limits of integration.

🎯 Exam Tip: Remember to convert \( \frac{1}{\sqrt{x}} \) to \( x^{- \frac{1}{2}} \) for easier integration and carefully apply the upper and lower limits.

Question 2. \( \int_{-2}^{3} \frac{1}{x+5} dx \)
Answer: Solution: \( \int_{-2}^{3} \frac{1}{x+5} dx \) \( = [ \log |x + 5| ]_{-2}^{3} \) \( = \log |3 + 5| - \log |-2 + 5| \) \( = \log 8 - \log 3 \) \( = \log \left(\frac{8}{3}\right) \) In simple words: The integral of \( \frac{1}{x} \) is \( \log|x| \). We apply this rule and then subtract the value at the lower limit from the value at the upper limit.

🎯 Exam Tip: Recall that the integral of \( \frac{1}{ax+b} \) is \( \frac{1}{a} \log|ax+b| \) and ensure correct application of limits for definite integrals.

Question 3. \( \int_{2}^{3} \frac{x}{x^2-1} dx \)
Answer: Solution: \( \int_{2}^{3} \frac{x}{x^2-1} dx \) \( = \frac{1}{2} \int_{2}^{3} \frac{2x}{x^2-1} dx \) \( = \frac{1}{2} [\log |x^2-1|]_{2}^{3} \dots \left[ \frac{d}{dx}(x^2-1) = 2x \text{ and } \int \frac{f'(x)}{f(x)} dx = \log|f(x)|+c \right] \) \( = \frac{1}{2} [\log (3^2-1) - \log (2^2-1)] \) \( = \frac{1}{2} [\log (9-1) - \log (4-1)] = \frac{1}{2} [\log 8 - \log 3] \) \( = \frac{1}{2} \log \left(\frac{8}{3}\right) \) In simple words: We manipulate the integrand to match the form \( \frac{f'(x)}{f(x)} \), whose integral is \( \log|f(x)| \), and then evaluate using the given limits.

🎯 Exam Tip: Look for opportunities to use the substitution method or the \( \int \frac{f'(x)}{f(x)} dx \) formula. Pay attention to the constant factor needed for the derivative.

Question 4. \( \int_{0}^{1} \frac{x^2+3x+2}{\sqrt{x}} dx \)
Answer: Solution: \( = \int_{0}^{1} \frac{x^2+3x+2}{\sqrt{x}} dx \) \( = \int_{0}^{1} \left( \frac{x^2}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{2}{\sqrt{x}} \right) dx \) \( = \int_{0}^{1} (x^{2 - \frac{1}{2}} + 3x^{1 - \frac{1}{2}} + 2x^{- \frac{1}{2}}) dx \) \( = \int_{0}^{1} (x^{\frac{3}{2}} + 3x^{\frac{1}{2}} + 2x^{- \frac{1}{2}}) dx \) \( = \left[ \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} + 3 \left( \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right) + 2 \left( \frac{x^{- \frac{1}{2}+1}}{- \frac{1}{2}+1} \right) \right]_{0}^{1} \) \( = \left[ \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + 3 \left( \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right) + 2 \left( \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right) \right]_{0}^{1} \) \( = \left[ \frac{2}{5} x^{\frac{5}{2}} + 2x^{\frac{3}{2}} + 4x^{\frac{1}{2}} \right]_{0}^{1} \) \( = \left[ \frac{2}{5} (1)^{\frac{5}{2}} + 2(1)^{\frac{3}{2}} + 4(1)^{\frac{1}{2}} \right] - \left[ \frac{2}{5} (0)^{\frac{5}{2}} + 2(0)^{\frac{3}{2}} + 4(0)^{\frac{1}{2}} \right] \) \( = \left[ \frac{2}{5} + 2 + 4 \right] - (0 + 0 + 0) \) \( = \frac{2}{5} + 6 = \frac{2 + 30}{5} = \frac{32}{5}. \) In simple words: We first divide each term in the numerator by \( \sqrt{x} \) and convert to fractional exponents, then integrate each term using the power rule, and finally apply the definite limits.

🎯 Exam Tip: Simplify the integrand by dividing and using exponent rules before integrating. Be careful with the lower limit of 0, as some terms might become zero.

Question 5. \( \int_{2}^{3} \frac{x}{(x+2)(x+3)} dx \)
Answer: Solution: Let \( I = \int_{2}^{3} \frac{x}{(x+2)(x+3)} dx \) Let \( \frac{x}{(x+2)(x+3)} = \frac{A}{x+3} + \frac{B}{x+2} \)
\( \implies x = A(x + 2) + B(x + 3) \) Put x + 3 = 0, i.e. x = -3, we get -3 = A(-3 + 2) + B(-3 + 3) -3 = A(-1) + B(0)
\( \implies A = 3 \) Put x + 2 = 0, i.e. x = -2, we get -2 = A(-2 + 2) + B(-2 + 3) -2 = A(0) + B(1)
\( \implies B = -2 \) Thus, \( \frac{x}{(x+2)(x+3)} = \frac{3}{x+3} + \frac{(-2)}{x+2} \)
\( \implies I = \int_{2}^{3} \left[ \frac{3}{x+3} - \frac{2}{x+2} \right] dx \) \( = [ 3 \log |x + 3| - 2\log |x+2| ]_{2}^{3} \) \( = [3\log (3+3)-2\log (3+2)] - [3\log (2+3)-2\log (2+2)] \) \( = [3\log 6 - 2\log 5] - [3\log 5 - 2\log 4] \) \( = 3\log 6 - 2\log 5 - 3\log 5 + 2\log 4 \) \( = 3\log 6 - 5\log 5 + 2\log 4 \) \( = \log 6^3 - \log 5^5 + \log 4^2 \) \( = \log 216 - \log 3125 + \log 16 \) \( = \log \left( \frac{216 \times 16}{3125} \right) = \log \left( \frac{3456}{3125} \right) \) In simple words: We decompose the integrand into partial fractions, integrate each simple fraction (which results in logarithmic terms), and then apply the limits of integration.

🎯 Exam Tip: Partial fraction decomposition is a key technique for integrating rational functions. Ensure you correctly find the constants A and B and handle the logarithmic properties at the end.

Question 6. \( \int_{1}^{2} \frac{dx}{x^2+6x+5} \)
Answer: Solution: \( \int_{1}^{2} \frac{dx}{x^2+6x+5} \) We complete the square in the denominator: \( x^2+6x+5 = (x^2+6x+9) - 9 + 5 = (x+3)^2 - 4 \) \( = \int_{1}^{2} \frac{dx}{(x+3)^2-4} \) \( = \int_{1}^{2} \frac{1}{(x+3)^2-(2)^2} dx \) Using the formula \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \) \( = \frac{1}{2(2)} \left[ \log \left| \frac{x+3-2}{x+3+2} \right| \right]_{1}^{2} \) \( = \frac{1}{4} \left[ \log \left| \frac{x+1}{x+5} \right| \right]_{1}^{2} \) \( = \frac{1}{4} \left[ \log \left| \frac{2+1}{2+5} \right| - \log \left| \frac{1+1}{1+5} \right| \right] \) \( = \frac{1}{4} \left[ \log \frac{3}{7} - \log \frac{2}{6} \right] \) \( = \frac{1}{4} \left[ \log \frac{3}{7} - \log \frac{1}{3} \right] \) \( = \frac{1}{4} \log \left( \frac{3/7}{1/3} \right) \) \( = \frac{1}{4} \log \left( \frac{3}{7} \times 3 \right) \) \( = \frac{1}{4} \log \left( \frac{9}{7} \right) \) In simple words: First, complete the square in the denominator to convert the quadratic expression into the form \( (x+a)^2 - b^2 \), then apply the standard integral formula for \( \frac{1}{x^2-a^2} \) and evaluate with the given limits.

🎯 Exam Tip: Recognizing standard integral forms like \( \int \frac{1}{x^2-a^2} dx \) after completing the square is crucial. Carefully handle the absolute value in the logarithm and properties of logarithms during simplification.

Question 7. If \( \int_{0}^{a} (2x + 1)dx = 2 \), find the real values of 'a'.
Answer: Solution: Let \( I = \int_{0}^{a} (2x + 1)dx \) \( = \left[ 2 \cdot \frac{x^2}{2} + x \right]_{0}^{a} \) \( = [ x^2 + x ]_{0}^{a} \) \( = (a^2 + a) - (0^2 + 0) \) \( = a^2 + a \)
\( \implies I = 2 \text{ gives } a^2 + a = 2 \)
\( \implies a^2 + a - 2 = 0 \)
\( \implies (a + 2)(a - 1) = 0 \)
\( \implies a + 2 = 0 \text{ or } a - 1 = 0 \)
\( \implies a = -2 \text{ or } a = 1. \) In simple words: We evaluate the definite integral in terms of 'a', set it equal to 2, and solve the resulting quadratic equation to find the possible values of 'a'.

🎯 Exam Tip: Integrate correctly and substitute the limits precisely. Solving the resulting algebraic equation (often quadratic or cubic) accurately is the final step.

Question 8. If \( \int_{1}^{a} (3x^2 + 2x + 1) dx = 11 \), find 'a'.
Answer: Solution: Let \( I = \int_{1}^{a} (3x^2 + 2x + 1) dx \) \( = \left[ 3 \left( \frac{x^3}{3} \right) + 2 \left( \frac{x^2}{2} \right) + x \right]_{1}^{a} \) \( = [ x^3 + x^2 + x ]_{1}^{a} \) \( = (a^3 + a^2 + a) - (1^3 + 1^2 + 1) \) \( = a^3 + a^2 + a - 3 \)
\( \implies I = 11 \text{ gives } a^3 + a^2 + a - 3 = 11 \)
\( \implies a^3 + a^2 + a - 14 = 0 \) By inspection, if \( a=2 \), then \( 2^3 + 2^2 + 2 - 14 = 8 + 4 + 2 - 14 = 0 \), so \( (a-2) \) is a factor. We can rewrite the equation to factor \( (a-2) \):
\( \implies (a^3 - 8) + (a^2 + a - 6) = 0 \)
\( \implies (a - 2)(a^2 + 2a + 4) + (a + 3)(a - 2) = 0 \)
\( \implies (a - 2)[(a^2 + 2a + 4) + (a + 3)] = 0 \)
\( \implies (a - 2)(a^2 + 3a + 7) = 0 \)
\( \implies a - 2 = 0 \text{ or } a^2 + 3a + 7 = 0 \)
\( \implies a = 2 \text{ or } a = \frac{-3 \pm \sqrt{3^2-4(1)(7)}}{2} = \frac{-3 \pm \sqrt{9-28}}{2} \) The latter two roots are not real because the discriminant \( (9-28 = -19) \) is negative.
\( \implies \text{they are rejected.} \)
\( \implies a = 2. \) In simple words: We integrate the polynomial, substitute the limits, and set the result equal to 11. This leads to a cubic equation which is solved by finding a real root through inspection and then factoring to check for other real roots.

🎯 Exam Tip: For cubic equations, try integer values near the constant term's divisors to find a root. Once a root (e.g., a=2) is found, use polynomial division to factorize the cubic and then solve the resulting quadratic equation.

Question 9. \( \int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} dx \)
Answer: Solution: Let \( I = \int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} dx \) Rationalize the denominator by multiplying by the conjugate: \( = \int_{0}^{1} \frac{1}{\sqrt{1+x}+\sqrt{x}} \times \frac{\sqrt{1+x}-\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} dx \) \( = \int_{0}^{1} \frac{\sqrt{1+x}-\sqrt{x}}{(\sqrt{1+x})^2-(\sqrt{x})^2} dx \) \( = \int_{0}^{1} \frac{\sqrt{1+x}-\sqrt{x}}{(1+x)-x} dx \) \( = \int_{0}^{1} \frac{\sqrt{1+x}-\sqrt{x}}{1} dx \) \( = \int_{0}^{1} ((1+x)^{\frac{1}{2}} - x^{\frac{1}{2}}) dx \) \( = \left[ \frac{(1+x)^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]_{0}^{1} - \left[ \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]_{0}^{1} \) \( = \left[ \frac{(1+x)^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{1} - \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{1} \) \( = \frac{2}{3} [(1+x)^{\frac{3}{2}}]_{0}^{1} - \frac{2}{3} [x^{\frac{3}{2}}]_{0}^{1} \) \( = \frac{2}{3} [(1+1)^{\frac{3}{2}} - (1+0)^{\frac{3}{2}}] - \frac{2}{3} [1^{\frac{3}{2}} - 0^{\frac{3}{2}}] \) \( = \frac{2}{3} [2^{\frac{3}{2}} - 1^{\frac{3}{2}}] - \frac{2}{3} [1 - 0] \) \( = \frac{2}{3} [2\sqrt{2} - 1] - \frac{2}{3} (1) \) \( = \frac{2}{3} (2\sqrt{2} - 1 - 1) \) \( = \frac{2}{3} (2\sqrt{2} - 2) \) \( = \frac{4}{3} (\sqrt{2} - 1). \) In simple words: Rationalize the denominator by multiplying by the conjugate, which simplifies the expression. Then, integrate each term separately using the power rule for integration and apply the definite limits.

🎯 Exam Tip: Rationalization is a key step when square roots are in the denominator. Simplify the expression before integrating to avoid errors. Be careful with fractional exponents and calculations during evaluation.

Question 10. \( \int_{1}^{2} \frac{3x}{9x^2-1} dx \)
Answer: Solution: Let \( I = \int_{1}^{2} \frac{3x}{9x^2-1} dx \) We can rewrite the denominator as \( (3x)^2-1^2 \). Method 1: Substitution Put \( 3x = t \)
\( \implies 3 dx = dt \)
\( \implies dx = \frac{dt}{3} \) Change the limits of integration: When \( x = 1, t = 3 \times 1 = 3 \) When \( x = 2, t = 3 \times 2 = 6 \)
\( \implies I = \int_{3}^{6} \frac{t}{t^2-1} \frac{dt}{3} = \frac{1}{3} \int_{3}^{6} \frac{t}{t^2-1} dt \) We need \( 2t \) in the numerator for \( \int \frac{f'(t)}{f(t)} dt \). \( = \frac{1}{3} \cdot \frac{1}{2} \int_{3}^{6} \frac{2t}{t^2-1} dt \) \( = \frac{1}{6} [ \log |t^2-1| ]_{3}^{6} \dots \left[ \frac{d}{dt} (t^2-1) = 2t \right] \) \( = \frac{1}{6} [ \log |6^2-1| - \log |3^2-1| ] \) \( = \frac{1}{6} [ \log |36-1| - \log |9-1| ] \) \( = \frac{1}{6} [ \log 35 - \log 8 ] \) \( = \frac{1}{6} \log \left( \frac{35}{8} \right) \) Alternative Method: Direct application of \( \int \frac{f'(x)}{f(x)} dx \) \( \int_{1}^{2} \frac{3x}{9x^2-1} dx \) The derivative of \( 9x^2-1 \) is \( 18x \). We have \( 3x \) in the numerator. \( = \frac{1}{6} \int_{1}^{2} \frac{18x}{9x^2-1} dx \) \( = \frac{1}{6} [ \log |9x^2-1| ]_{1}^{2} \dots \left[ \frac{d}{dx} (9x^2-1) = 18x \text{ and } \int \frac{f'(x)}{f(x)} dx = \log |f(x)| \right] \) \( = \frac{1}{6} [ \log |9(2)^2-1| - \log |9(1)^2-1| ] \) \( = \frac{1}{6} [ \log |9 \times 4 - 1| - \log |9 \times 1 - 1| ] \) \( = \frac{1}{6} [ \log |36 - 1| - \log |9 - 1| ] \) \( = \frac{1}{6} [ \log 35 - \log 8 ] \) \( = \frac{1}{6} \log \left( \frac{35}{8} \right) \) In simple words: We can solve this integral either by using a substitution (let \( t = 3x \) or \( t = 9x^2-1 \)) or by directly adjusting the numerator to be the derivative of the denominator, then integrating using the logarithmic form \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| \), and applying the limits.

🎯 Exam Tip: For integrals of the form \( \frac{ax}{bx^2+c} \), always check if the numerator can be made proportional to the derivative of the denominator to use the logarithmic integration rule. Substitution is another robust approach.

Question 11. \( \int_{1}^{3} \log x dx \)
Answer: Solution: Let \( I = \int_{1}^{3} \log x dx \) We use integration by parts, \( \int u dv = uv - \int v du \). Let \( u = \log x \) and \( dv = 1 dx \). Then \( du = \frac{1}{x} dx \) and \( v = x \). \( = [ (\log x) \cdot x ]_{1}^{3} - \int_{1}^{3} x \cdot \frac{1}{x} dx \) \( = [ x \log x ]_{1}^{3} - \int_{1}^{3} 1 dx \) \( = [ (3 \log 3) - (1 \log 1) ] - [x]_{1}^{3} \dots [\dots \log 1 = 0] \) \( = 3 \log 3 - 0 - (3 - 1) \) \( = 3 \log 3 - 2 \) In simple words: We solve this integral using integration by parts, treating \( \log x \) as \( u \) and \( 1 \) as \( dv \). After applying the formula, we evaluate the definite integral by substituting the upper and lower limits.

🎯 Exam Tip: Remember the LIATE rule for choosing \( u \) in integration by parts. For \( \int \log x dx \), it's a common trick to let \( dv = 1 dx \). Be careful with the evaluation of \( \log 1 \).