CBSE Class 12 Maths: Integrals Exercise 7.10 (Definite Integrals by Substitution)

Understanding Definite Integrals and the Substitution Method

Definite integrals, unlike indefinite integrals, evaluate to a specific numerical value. They represent the area under a curve between two fixed points, known as the lower and upper limits of integration. While indefinite integration gives a family of functions, definite integration provides a precise quantity. The Fundamental Theorem of Calculus links these two concepts, stating that if F(x) is an antiderivative of f(x), then \(\int_a^b f(x) dx = F(b) - F(a)\).<br/><br/>Exercise 7.10 primarily focuses on solving definite integrals using the method of substitution. This method is invaluable when the integrand (the function being integrated) is a composite function, or when it contains a function and its derivative. When applying substitution to definite integrals, a critical step is to change the limits of integration according to the new variable. If we substitute \(u = g(x)\), and the original limits for \(x\) are \(a\) and \(b\), then the new limits for \(u\) will be \(g(a)\) and \(g(b)\) respectively. Failing to change the limits correctly is a very common mistake. Once the substitution is made and limits are changed, the integral becomes simpler to evaluate in terms of the new variable, \(u\), without needing to revert back to \(x\) at the end.

Key Definitions and Properties for Exercise 7.10

Definite Integral

An integral \(\int_a^b f(x) dx\) that represents the area under the curve \(y = f(x)\) from \(x = a\) to \(x = b\). Its value is a single number, calculated as \(F(b) - F(a)\) where \(F(x)\) is an antiderivative of \(f(x)\).

Substitution Method for Definite Integrals

A technique to simplify integrals by introducing a new variable, \(u = g(x)\). For definite integrals, it's crucial to change the limits of integration: if \(x\) goes from \(a\) to \(b\), then \(u\) goes from \(g(a)\) to \(g(b)\).

Property P3 (King Property)

This property states that \(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\). It is extremely useful for simplifying integrals where the sum of limits simplifies the integrand, often used when dealing with trigonometric or logarithmic functions.

Property P2 (Interval Splitting)

For any \(c\) such that \(a < c < b\), we have \(\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx\). This property is useful when the function \(f(x)\) has different definitions in different intervals or involves absolute value functions.

Worked Examples: Definite Integrals by Substitution and Properties

• Example 1: Evaluate \(\int_0^1 \frac{x}{x^2 + 1} dx\) Step 1: Identify substitution. Let \(t = x^2 + 1\). Step 2: Find \(dt/dx\). \(\frac{dt}{dx} = 2x \Rightarrow dx = \frac{dt}{2x}\). Step 3: Change limits of integration. When \(x = 0\), \(t = 0^2 + 1 = 1\). When \(x = 1\), \(t = 1^2 + 1 = 2\). Step 4: Substitute into the integral. \(\int_1^2 \frac{x}{t} \cdot \frac{dt}{2x} = \int_1^2 \frac{1}{2t} dt\) Step 5: Integrate with respect to \(t\). \(\frac{1}{2} [\ln|t|]_1^2\) Step 6: Apply the limits. \(\frac{1}{2} (\ln 2 - \ln 1) = \frac{1}{2} (\ln 2 - 0) = \frac{1}{2} \ln 2\) Final answer: \(\frac{1}{2} \ln 2\)
• Example 2: Evaluate \(\int_0^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} dx\) Step 1: Let the integral be \(I\). \(I = \int_0^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} dx\) (Equation 1) Step 2: Apply Property P3: \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\). Here, \(a = \pi/2\). So, \(x\) becomes \(\pi/2 - x\). \(\sin(\pi/2 - x) = \cos x\) and \(\cos(\pi/2 - x) = \sin x\). \(I = \int_0^{\pi/2} \frac{\sin^4 (\pi/2 - x)}{\sin^4 (\pi/2 - x) + \cos^4 (\pi/2 - x)} dx\) \(I = \int_0^{\pi/2} \frac{\cos^4 x}{\cos^4 x + \sin^4 x} dx\) (Equation 2) Step 3: Add Equation 1 and Equation 2. \(I + I = \int_0^{\pi/2} \left( \frac{\sin^4 x}{\sin^4 x + \cos^4 x} + \frac{\cos^4 x}{\cos^4 x + \sin^4 x} \right) dx\) \(2I = \int_0^{\pi/2} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} dx\) \(2I = \int_0^{\pi/2} 1 dx\) Step 4: Integrate and apply limits. \(2I = [x]_0^{\pi/2}\) \(2I = \pi/2 - 0\) \(2I = \pi/2\) Step 5: Solve for \(I\). \(I = \frac{\pi}{4}\) Final answer: \(\frac{\pi}{4}\)

Exam Tips for Definite Integrals (Exercise 7.10)

1. Don't Forget to Change Limits: This is the single most common mistake in definite integration by substitution. When you change the variable from \(x\) to \(t\) (or \(u\)), you must also change the limits from \(x_1, x_2\) to \(t_1 = g(x_1)\) and \(t_2 = g(x_2)\). If you don't change the limits, you must revert back to the original variable after integration before applying the original limits, which is often more cumbersome.
2. Identify the Correct Substitution: Look for a function and its derivative within the integrand. For example, if you see \(f'(x) \cdot f(x)^n\), substitute \(t = f(x)\). If you see \(f'(x) / f(x)\), substitute \(t = f(x)\).
3. Master the Properties: Particularly the 'King Property' (\(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\)) and its variant (\(\int_0^a f(x) dx = \int_0^a f(a-x) dx\)). These properties can transform seemingly complex integrals into incredibly simple ones (as seen in Example 2).
4. Simplify Before Integrating: Sometimes, algebraic manipulation or trigonometric identities can simplify the integrand before you even think about substitution or properties. Always look for simplification opportunities.
5. Practice is Key: The more problems you solve, the better you become at identifying the right substitution or property to apply. Work through all problems in Exercise 7.10 and similar examples from your reference books.

Practice Questions with Solutions

• Q: Evaluate \(\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx\) A: Step 1: Let \(I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx\) (Eq. 1) Step 2: Apply the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\). Here \(a = \pi/2\). \(I = \int_0^{\pi/2} \frac{\sqrt{\sin (\pi/2 - x)}}{\sqrt{\sin (\pi/2 - x)} + \sqrt{\cos (\pi/2 - x)}} dx\) \(I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\) (Eq. 2) Step 3: Add Eq. 1 and Eq. 2. \(2I = \int_0^{\pi/2} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \right) dx\) \(2I = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx\) \(2I = \int_0^{\pi/2} 1 dx\) Step 4: Integrate and apply limits. \(2I = [x]_0^{\pi/2}\) \(2I = \pi/2 - 0\) \(2I = \pi/2\) Step 5: Solve for \(I\). \(I = \frac{\pi}{4}\) Final answer: \(\frac{\pi}{4}\)
• Q: Evaluate \(\int_0^1 \frac{\tan^{-1} x}{1 + x^2} dx\) A: Step 1: Identify substitution. Let \(t = \tan^{-1} x\). Step 2: Find \(dt/dx\). \(\frac{dt}{dx} = \frac{1}{1 + x^2} \Rightarrow dt = \frac{1}{1 + x^2} dx\). Step 3: Change limits of integration. When \(x = 0\), \(t = \tan^{-1} 0 = 0\). When \(x = 1\), \(t = \tan^{-1} 1 = \pi/4\). Step 4: Substitute into the integral. \(\int_0^{\pi/4} t dt\) Step 5: Integrate with respect to \(t\). \(\left[ \frac{t^2}{2} \right]_0^{\pi/4}\) Step 6: Apply the limits. \(\frac{1}{2} \left( (\pi/4)^2 - 0^2 \right) = \frac{1}{2} \left( \frac{\pi^2}{16} \right) = \frac{\pi^2}{32}\) Final answer: \(\frac{\pi^2}{32}\)
• Q: Evaluate \(\int_0^2 x \sqrt{x+2} dx\) A: Step 1: Identify substitution. Let \(t = x+2\). Then \(x = t-2\) and \(dx = dt\). Step 2: Change limits of integration. When \(x = 0\), \(t = 0+2 = 2\). When \(x = 2\), \(t = 2+2 = 4\). Step 3: Substitute into the integral. \(\int_2^4 (t-2) \sqrt{t} dt = \int_2^4 (t^{3/2} - 2t^{1/2}) dt\) Step 4: Integrate with respect to \(t\). \(\left[ \frac{t^{5/2}}{5/2} - 2 \frac{t^{3/2}}{3/2} \right]_2^4 = \left[ \frac{2}{5} t^{5/2} - \frac{4}{3} t^{3/2} \right]_2^4\) Step 5: Apply the limits. \(\left( \frac{2}{5} (4)^{5/2} - \frac{4}{3} (4)^{3/2} \right) - \left( \frac{2}{5} (2)^{5/2} - \frac{4}{3} (2)^{3/2} \right)\) \(= \left( \frac{2}{5} (32) - \frac{4}{3} (8) \right) - \left( \frac{2}{5} (4\sqrt{2}) - \frac{4}{3} (2\sqrt{2}) \right)\) \(= \left( \frac{64}{5} - \frac{32}{3} \right) - \left( \frac{8\sqrt{2}}{5} - \frac{8\sqrt{2}}{3} \right)\) \(= \left( \frac{192 - 160}{15} \right) - \left( \frac{24\sqrt{2} - 40\sqrt{2}}{15} \right)\) \(= \frac{32}{15} - \left( \frac{-16\sqrt{2}}{15} \right) = \frac{32 + 16\sqrt{2}}{15}\) Final answer: \(\frac{32 + 16\sqrt{2}}{15}\)
• Q: Evaluate \(\int_{-1}^1 \frac{1}{x^2 + 2x + 5} dx\) A: Step 1: Complete the square in the denominator. \(x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2\). The integral becomes \(\int_{-1}^1 \frac{1}{(x+1)^2 + 2^2} dx\) Step 2: Identify substitution. Let \(t = x+1\). Then \(dt = dx\). Step 3: Change limits of integration. When \(x = -1\), \(t = -1+1 = 0\). When \(x = 1\), \(t = 1+1 = 2\). Step 4: Substitute into the integral. \(\int_0^2 \frac{1}{t^2 + 2^2} dt\) Step 5: Integrate using the formula \(\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C\). Here \(a=2\), so \(\left[ \frac{1}{2} \tan^{-1} \left( \frac{t}{2} \right) \right]_0^2\) Step 6: Apply the limits. \(\frac{1}{2} \left( \tan^{-1} \left( \frac{2}{2} \right) - \tan^{-1} \left( \frac{0}{2} \right) \right)\) \(= \frac{1}{2} (\tan^{-1} 1 - \tan^{-1} 0)\) \(= \frac{1}{2} (\pi/4 - 0) = \frac{\pi}{8}\) Final answer: \(\frac{\pi}{8}\)

Frequently Asked Questions