Integrals Ex 7.11 Class 12 NCERT: Properties of Definite Integrals

Understanding the Properties of Definite Integrals

Definite integration measures the net signed area under a curve. While the Fundamental Theorem of Calculus allows us to evaluate these by finding the anti-derivative $F(x)$, many functions are practically impossible to integrate directly. This is where the properties of definite integrals step in. By exploiting the symmetry, periodic nature, or boundaries of the integrand, we can bypass complex anti-differentiation entirely.

For instance, the split-interval property allows us to evaluate piecewise functions like $|x|$ by breaking the integral at points of discontinuity. The 'King's Property' ($\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$) helps eliminate complex trigonometric denominators. Using these rules strategically is the key to cracking Ex 7.11 efficiently.

Mathematical Formulation of Key Properties

P3: Splitting Property

$\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$, where $c \in (a, b)$. Mostly used for modulus and step functions.

P4: King's Property

$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. This is highly effective when the sum of limits simplifies the expression.

P5: Special King's Property

$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$. A special case of P4 where the lower limit is zero.

P7: Even and Odd Functions

$\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(-x) = f(x)$ (even), and $0$ if $f(-x) = -f(x)$ (odd).

Systematic Approach to Solve Definite Integrals Using Properties

1. Identify the Integrand Type — Determine if the function contains modulus symbols, symmetric limits (like $-a$ to $a$), or trigonometric ratios that might simplify using $(a-x)$.
2. Apply the Appropriate Property — For symmetric limits $[-a, a]$, test if the function is even or odd. For modulus, split at the critical points where the expression inside the mod becomes zero. For trigonometric terms, apply the King's property.
3. Combine Equations (if using P4/P5) — Let the original integral be $I$. Apply the property to get a new expression for $I$. Add the two integral equations ($2I = \int [f(x) + f(a-x)] dx$) to simplify the integrand.
4. Integrate and Evaluate Limits — Solve the simplified integrand (which often reduces to a constant or 1) and substitute the upper and lower limits to find the final value.

CBSE Board Exam Tips & Common Traps

1. Modulus Sign Changes: Always explicitly state the interval breakdown when splitting modulus integrals. For example, for $\int_{-1}^{2} |x| dx$, show that $|x| = -x$ for $[-1, 0)$ and $+x$ for $[0, 2]$.

1. The $2I$ Trap: A very common error in applying P4/P5 is forgetting that adding the two equations yields $2I$, not $I$. Remember to divide the final integrated value by 2 to get the actual value of $I$.

1. Odd Function Checklist: Do not blindly write 0 for limits $[-a, a]$ without proving $f(-x) = -f(x)$. Show this step clearly in your board sheets to secure full marks.

Practice Questions with Solutions

• Q: Evaluate $\int_{0}^{\pi/2} \frac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x} dx$. A: Step 1: Let $I = \int_{0}^{\pi/2} \frac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x} dx$ ---(Equation 1) Step 2: Apply the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$. Replace $x$ with $(\pi/2 - x)$: $I = \int_{0}^{\pi/2} \frac{\sin^{3/2}(\pi/2 - x)}{\sin^{3/2}(\pi/2 - x) + \cos^{3/2}(\pi/2 - x)} dx$ Since $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$, we get: $I = \int_{0}^{\pi/2} \frac{\cos^{3/2}x}{\cos^{3/2}x + \sin^{3/2}x} dx$ ---(Equation 2) Step 3: Add Equation 1 and Equation 2: $2I = \int_{0}^{\pi/2} \frac{\sin^{3/2}x + \cos^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x} dx$ $2I = \int_{0}^{\pi/2} 1 \cdot dx$ Step 4: Integrate and evaluate: $2I = [x]_{0}^{\pi/2} = \pi/2 - 0 = \pi/2$ $I = \pi/4$ Final answer: $\pi/4$
• Q: Evaluate $\int_{-5}^{5} |x + 2| dx$. A: Step 1: Find the critical point where the modulus inside changes sign. Set $x + 2 = 0 \implies x = -2$. Since $-2$ lies in the interval $[-5, 5]$, split the integral: $\int_{-5}^{5} |x + 2| dx = \int_{-5}^{-2} |x + 2| dx + \int_{-2}^{5} |x + 2| dx$ Step 2: Define the sign of $(x + 2)$ in each sub-interval: For $x \in [-5, -2]$, $|x + 2| = -(x + 2)$ For $x \in [-2, 5]$, $|x + 2| = (x + 2)$ Step 3: Substitute and integrate: $I = \int_{-5}^{-2} -(x + 2) dx + \int_{-2}^{5} (x + 2) dx$ $I = -[\frac{x^2}{2} + 2x]_{-5}^{-2} + [\frac{x^2}{2} + 2x]_{-2}^{5}$ Step 4: Evaluate the limits: First part: $-[(\frac{4}{2} - 4) - (\frac{25}{2} - 10)] = -[-2 - 2.5] = 4.5$ Second part: $[(\frac{25}{2} + 10) - (\frac{4}{2} - 4)] = [22.5 - (-2)] = 24.5$ Total $I = 4.5 + 24.5 = 29$ Final answer: 29
• Q: Evaluate $\int_{-\pi/2}^{\pi/2} \sin^7 x dx$. A: Step 1: Identify symmetric limits $[-a, a]$ where $a = \pi/2$. We must test if $f(x) = \sin^7 x$ is even or odd. Step 2: Substitute $-x$ for $x$: $f(-x) = \sin^7(-x) = (-\sin x)^7 = -\sin^7 x = -f(x)$ Since $f(-x) = -f(x)$, the function is odd. Step 3: Apply the symmetric limits property (P7): $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is odd. Therefore, $\int_{-\pi/2}^{\pi/2} \sin^7 x dx = 0$ Final answer: 0
• Q: Evaluate $\int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx$. A: Step 1: Let $I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx$ ---(Equation 1) Step 2: Apply property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$. Replace $x$ with $(\pi - x)$: $I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx$ Since $\sin(\pi - x) = \sin x$ and $\cos^2(\pi - x) = \cos^2 x$, we get: $I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx$ ---(Equation 2) Step 3: Add Equation 1 and Equation 2: $2I = \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} dx \implies I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$ Step 4: Use substitution $t = \cos x \implies dt = -\sin x dx$. When $x = 0, t = 1$; when $x = \pi, t = -1$. $I = \frac{\pi}{2} \int_{1}^{-1} \frac{-dt}{1 + t^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{dt}{1 + t^2}$ Since $1/(1+t^2)$ is even: $I = \frac{\pi}{2} \cdot 2 \int_{0}^{1} \frac{dt}{1 + t^2} = \pi [\tan^{-1} t]_{0}^{1} = \pi (\pi/4 - 0) = \frac{\pi^2}{4}$ Final answer: $\frac{\pi^2}{4}$

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