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Integrals - Some Properties of Definite Integrals

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The value of a definite integral does not change if the variable of integration is changed, provided the limits remain the same. This is expressed as ∫abf(x)dx=∫abf(t)dt\int_{a}^{b} f(x) dx = \int_{a}^{b} f(t) dt.

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Interchanging the limits of integration changes the sign of the definite integral: ∫abf(x)dx=−∫baf(x)dx\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx.

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The integral over an interval [a,b][a, b] can be split into the sum of integrals over sub-intervals [a,c][a, c] and [c,b][c, b]. This is particularly useful for piecewise or modulus functions.

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The property ∫0af(x)dx=∫0af(a−x)dx\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx is often called the 'King's Rule' and is extremely effective for simplifying trigonometric integrals.

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For symmetric limits [−a,a][-a, a], if f(x)f(x) is an odd function (i.e., f(−x)=−f(x)f(-x) = -f(x)), the integral is 00. If f(x)f(x) is an even function (i.e., f(−x)=f(x)f(-x) = f(x)), the integral is 2∫0af(x)dx2 \int_{0}^{a} f(x) dx.

📐Formulae

∫abf(x)dx=∫abf(t)dt\int_{a}^{b} f(x) dx = \int_{a}^{b} f(t) dt

∫abf(x)dx=−∫baf(x)dx\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx

∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx

∫abf(x)dx=∫abf(a+b−x)dx\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx

∫0af(x)dx=∫0af(a−x)dx\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx

∫−aaf(x)dx={2∫0af(x)dx,if f(−x)=f(x)0,if f(−x)=−f(x)\int_{-a}^{a} f(x) dx = \begin{cases} 2\int_{0}^{a} f(x) dx, & \text{if } f(-x) = f(x) \\ 0, & \text{if } f(-x) = -f(x) \end{cases}

∫02af(x)dx={2∫0af(x)dx,if f(2a−x)=f(x)0,if f(2a−x)=−f(x)\int_{0}^{2a} f(x) dx = \begin{cases} 2\int_{0}^{a} f(x) dx, & \text{if } f(2a-x) = f(x) \\ 0, & \text{if } f(2a-x) = -f(x) \end{cases}

💡Examples

Problem 1:

Evaluate I=∫0π/2sin⁡3/2xsin⁡3/2x+cos⁡3/2xdxI = \int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx.

Solution:

Let I=∫0π/2sin⁡3/2xsin⁡3/2x+cos⁡3/2xdx…(1)I = \int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx \dots (1). Using the property ∫0af(x)dx=∫0af(a−x)dx\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx, we get: I=∫0π/2sin⁡3/2(π/2−x)sin⁡3/2(π/2−x)+cos⁡3/2(π/2−x)dxI = \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⁡(π/2−x)=cos⁡x\sin(\pi/2 - x) = \cos x and cos⁡(π/2−x)=sin⁡x\cos(\pi/2 - x) = \sin x, we have I=∫0π/2cos⁡3/2xcos⁡3/2x+sin⁡3/2xdx…(2)I = \int_{0}^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx \dots (2). Adding (1) and (2): 2I=∫0π/2sin⁡3/2x+cos⁡3/2xsin⁡3/2x+cos⁡3/2xdx=∫0π/21dx2I = \int_{0}^{\pi/2} \frac{\sin^{3/2} x + \cos^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx = \int_{0}^{\pi/2} 1 dx. Thus 2I=[x]0π/2=π/22I = [x]_{0}^{\pi/2} = \pi/2, which gives I=π/4I = \pi/4.

Explanation:

This example applies the 'King's Rule' (P4P_4). By replacing xx with (π/2−x)(\pi/2 - x), the denominator remains the same while the numerator switches from sine to cosine. Adding the two forms eliminates the variable terms.

Problem 2:

Evaluate ∫25∣x−3∣dx\int_{2}^{5} |x-3| dx.

Solution:

The function ∣x−3∣|x-3| changes definition at x=3x=3. Within [2,5][2, 5], we split the integral at x=3x=3: ∫25∣x−3∣dx=∫23∣x−3∣dx+∫35∣x−3∣dx\int_{2}^{5} |x-3| dx = \int_{2}^{3} |x-3| dx + \int_{3}^{5} |x-3| dx. For x∈[2,3]x \in [2, 3], ∣x−3∣=−(x−3)=3−x|x-3| = -(x-3) = 3-x. For x∈[3,5]x \in [3, 5], ∣x−3∣=x−3|x-3| = x-3. Therefore: I=∫23(3−x)dx+∫35(x−3)dxI = \int_{2}^{3} (3-x) dx + \int_{3}^{5} (x-3) dx. I=[3x−x22]23+[x22−3x]35I = [3x - \frac{x^2}{2}]_{2}^{3} + [\frac{x^2}{2} - 3x]_{3}^{5}. I=(9−4.5)−(6−2)+(12.5−15)−(4.5−9)=0.5+2=2.5I = (9 - 4.5) - (6 - 2) + (12.5 - 15) - (4.5 - 9) = 0.5 + 2 = 2.5.

Explanation:

This demonstrates the splitting property (P2P_2). Since the modulus function is piecewise, the integral must be broken down based on the sign of the expression inside the absolute value.

Problem 3:

Evaluate ∫−π/2π/2sin⁡7xdx\int_{-\pi/2}^{\pi/2} \sin^7 x dx.

Solution:

Let f(x)=sin⁡7xf(x) = \sin^7 x. Then f(−x)=sin⁡7(−x)=(−sin⁡x)7=−sin⁡7xf(-x) = \sin^7(-x) = (-\sin x)^7 = -\sin^7 x. Since f(−x)=−f(x)f(-x) = -f(x), f(x)f(x) is an odd function. Using the property ∫−aaf(x)dx=0\int_{-a}^{a} f(x) dx = 0 for odd functions, we get ∫−π/2π/2sin⁡7xdx=0\int_{-\pi/2}^{\pi/2} \sin^7 x dx = 0.

Explanation:

This uses property P7P_7. Recognizing that a function is odd on a symmetric interval [−a,a][-a, a] allows for an immediate solution without performing complex integration.