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7.3 Definite Integrals, Properties & Fundamental Theorem of Calculus

Key Takeaways

  • The Fundamental Theorem of Calculus (FTC) bridges differential and integral calculus, asserting that if f(x)f(x) is continuous on [a,b][a, b] and F′(x)=f(x)F'(x) = f(x), then ∫abf(x)dx=F(b)−F(a)\int_a^b f(x) dx = F(b) - F(a).

  • King's Property ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) dx = \int_a^b f(a + b - x) dx and its canonical origin case ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a - x) dx enable instantaneous evaluation of symmetric quotients on the ECAT, yielding ∫0π/2sin⁡nxsin⁡nx+cos⁡nxdx=π4\int_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx = \frac{\pi}{4} for all n∈Rn \in \mathbb{R}.

  • For symmetric intervals [−a,a][-a, a], odd functions vanish identically (∫−aaf(x)dx=0\int_{-a}^a f(x) dx = 0), while even functions double over the half-interval (∫−aaf(x)dx=2∫0af(x)dx\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx).

  • Periodic functions of fundamental period TT satisfy ∫0nTf(x)dx=n∫0Tf(x)dx\int_0^{nT} f(x) dx = n \int_0^T f(x) dx and ∫aa+Tf(x)dx=∫0Tf(x)dx\int_a^{a + T} f(x) dx = \int_0^T f(x) dx, rendering the integral independent of the lower translational limit aa.

  • The Leibniz Rule for differentiating under the integral sign evaluates derivatives of variable-limit integrals without evaluating the antiderivative: ddx[∫u(x)v(x)f(t)dt]=f(v(x))v′(x)−f(u(x))u′(x)\frac{d}{dx}\left[\int_{u(x)}^{v(x)} f(t) dt\right] = f(v(x))v'(x) - f(u(x))u'(x).

Last updated: October 2026

Definite Integrals, Properties & Fundamental Theorem of Calculus

While indefinite integration determines an infinite family of primitive functions, the definite integral evaluates a unique numerical scalar representing the net accumulation of a quantity over a specified interval. On the Pakistan ECAT, definite integrals represent a high-yield topic tested through conceptual properties rather than cumbersome manual antiderivative computations. Familiarity with fundamental integral transformations—especially King's Property, parity symmetries, and periodic scaling—enables candidates to solve seemingly intractable definite integrals in under 15 seconds. This section presents the theoretical foundation of the definite integral, establishes the Fundamental Theorem of Calculus, and comprehensively derives the operational properties essential for test day.


1. The Fundamental Theorem of Calculus (FTC)

Historical and Conceptual Significance

Prior to the late seventeenth century, the calculation of slopes of tangent lines (differentiation) and the determination of planar areas bounded by curves (integration) were viewed as entirely disconnected mathematical problems. The Fundamental Theorem of Calculus, established independently by Isaac Newton and Gottfried Wilhelm Leibniz, proved that differentiation and integration are inverse mathematical processes.

First Fundamental Theorem of Calculus (Differentiability of Area Function)

Let ff be a continuous real-valued function on the closed interval [a,b][a, b], and define the accumulator function g(x)g(x) by:

g(x)=∫axf(t)dtfor x∈[a,b]g(x) = \int_a^x f(t) dt \quad \text{for } x \in [a, b]

Then g(x)g(x) is continuous on [a,b][a, b], differentiable on the open interval (a,b)(a, b), and its derivative is identically f(x)f(x):

g′(x)=ddx[∫axf(t)dt]=f(x)g'(x) = \frac{d}{dx}\left[ \int_a^x f(t) dt \right] = f(x)

Second Fundamental Theorem of Calculus (Evaluation Theorem)

If ff is continuous on [a,b][a, b] and FF is any antiderivative of ff on [a,b][a, b] (meaning F′(x)=f(x)F'(x) = f(x) for all x∈[a,b]x \in [a, b]), then:

∫abf(x)dx=[F(x)]ab=F(b)−F(a)\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)

Dummy Variable Invariance

Because the definite integral evaluates to a numerical value depending solely on the limits aa and bb and the functional form ff, the variable of integration is purely a dummy variable:

∫abf(x)dx=∫abf(t)dt=∫abf(u)du=∫abf(θ)dθ\int_a^b f(x) dx = \int_a^b f(t) dt = \int_a^b f(u) du = \int_a^b f(\theta) d\theta


2. Elementary Operational Properties of Definite Integrals

The following fundamental algebraic properties govern all definite integration:

1. Limits Reversal Property

Interchanging the limits of integration inverts the algebraic sign: ∫abf(x)dx=−∫baf(x)dx\int_a^b f(x) dx = -\int_b^a f(x) dx

2. Coincident Limits Property

An integral evaluated over an interval of zero width is identically zero: ∫aaf(x)dx=0\int_a^a f(x) dx = 0

3. Linearity of Definite Integrals

For any real constants α,β∈R\alpha, \beta \in \mathbb{R}: ∫ab[αf(x)±βg(x)]dx=α∫abf(x)dx±β∫abg(x)dx\int_a^b [\alpha f(x) \pm \beta g(x)] dx = \alpha \int_a^b f(x) dx \pm \beta \int_a^b g(x) dx

4. Interval Additivity (Splitting Property)

For any real number cc lying inside or outside the interval of integration (provided ff is integrable on the entire span): ∫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

Critical Application: Interval splitting is mandatory when evaluating integrals containing absolute values or piecewise-defined definitions:

Example: Evaluate ∫−23∣x∣dx\int_{-2}^3 |x| dx. Split at the critical point x=0x = 0 where ∣x∣|x| changes analytical definition: ∫−23∣x∣dx=∫−20(−x)dx+∫03xdx=[−x22]−20+[x22]03=(0−(−42))+(92−0)=2+4.5=6.5=132\int_{-2}^3 |x| dx = \int_{-2}^0 (-x) dx + \int_0^3 x dx = \left[ -\frac{x^2}{2} \right]_{-2}^0 + \left[ \frac{x^2}{2} \right]_0^3 = \left(0 - \left(-\frac{4}{2}\right)\right) + \left(\frac{9}{2} - 0\right) = 2 + 4.5 = 6.5 = \frac{13}{2}


3. King's Property: The ECAT Master Weapon

Among all definite integral identities, King's Property is the most powerful tool for solving complex trigonometric, rational, and logarithmic entrance exam questions.

Theorem Statement (General King's Property)

∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) dx = \int_a^b f(a + b - x) dx

Analytical Proof: Let t=a+b−xt = a + b - x. Differentiating gives dt=−dx  ⟹  dx=−dtdt = -dx \implies dx = -dt. Examine the transformed limits:

  • When x=a  ⟹  t=a+b−a=bx = a \implies t = a + b - a = b.
  • When x=b  ⟹  t=a+b−b=ax = b \implies t = a + b - b = a.

Substituting into the integral: ∫abf(x)dx=∫baf(a+b−t)(−dt)=−∫baf(a+b−t)dt=∫abf(a+b−t)dt\int_a^b f(x) dx = \int_b^a f(a + b - t) (-dt) = -\int_b^a f(a + b - t) dt = \int_a^b f(a + b - t) dt Since the integration variable is dummy, replace tt with xx: ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) dx = \int_a^b f(a + b - x) dx

Special Case: Integration from 00 to aa

Setting the lower limit a=0a = 0 yields the most frequently tested variant:

∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a - x) dx

The Universal Symmetric Quotient Theorem

A prominent class of ECAT questions asks for the definite integral of a symmetric fraction:

I=∫abf(x)f(x)+f(a+b−x)dxI = \int_a^b \frac{f(x)}{f(x) + f(a + b - x)} dx

Applying King's Property replaces xx with a+b−xa + b - x:

I=∫abf(a+b−x)f(a+b−x)+f(x)dxI = \int_a^b \frac{f(a + b - x)}{f(a + b - x) + f(x)} dx

Adding these two expressions eliminates the function entirely:

2I=∫abf(x)+f(a+b−x)f(x)+f(a+b−x)dx=∫ab1dx=b−a  ⟹  I=b−a22I = \int_a^b \frac{f(x) + f(a + b - x)}{f(x) + f(a + b - x)} dx = \int_a^b 1 dx = b - a \implies I = \frac{b - a}{2}

Canonical ECAT Results Deriving from King's Property

Candidates should memorize these three classic results:

  1. The Universal Sine-Cosine Power Integral: ∫0π/2sin⁡nxsin⁡nx+cos⁡nxdx=π/2−02=π4\int_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx = \frac{\pi/2 - 0}{2} = \frac{\pi}{4} This holds identically for all real values of nn (whether n=1,2,2024,12,3n = 1, 2, 2024, \frac{1}{2}, \sqrt{3}, or −5-5!).

  2. Tangent-Secant Equivalence: ∫0π/211+tan⁡nxdx=∫0π/2cos⁡nxcos⁡nx+sin⁡nxdx=π4\int_0^{\pi/2} \frac{1}{1 + \tan^n x} dx = \int_0^{\pi/2} \frac{\cos^n x}{\cos^n x + \sin^n x} dx = \frac{\pi}{4}

  3. The Famous Log-Sine Integral: ∫0π/2ln⁡(sin⁡x)dx=∫0π/2ln⁡(cos⁡x)dx=−π2ln⁡2\int_0^{\pi/2} \ln(\sin x) dx = \int_0^{\pi/2} \ln(\cos x) dx = -\frac{\pi}{2}\ln 2


4. Symmetry Properties (Even and Odd Functions)

When the interval of integration is symmetric about the origin, [−a,a][-a, a], evaluating the function's algebraic parity eliminates the need for integration entirely.

Mathematical Formulation

∫−aaf(x)dx={0if f(x) is an odd function (f(−x)=−f(x))2∫0af(x)dxif f(x) is an even function (f(−x)=f(x))\int_{-a}^a f(x) dx = \begin{cases} 0 & \text{if } f(x) \text{ is an odd function } (f(-x) = -f(x)) \\ 2 \int_0^a f(x) dx & \text{if } f(x) \text{ is an even function } (f(-x) = f(x)) \end{cases}

Proof for Odd Function: Split the integral at 00: ∫−aaf(x)dx=∫−a0f(x)dx+∫0af(x)dx\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx In the first integral, let x=−t  ⟹  dx=−dtx = -t \implies dx = -dt. The limits change from x=−a  ⟹  t=ax = -a \implies t = a to x=0  ⟹  t=0x = 0 \implies t = 0: ∫−a0f(x)dx=∫a0f(−t)(−dt)=∫0af(−t)dt\int_{-a}^0 f(x) dx = \int_a^0 f(-t) (-dt) = \int_0^a f(-t) dt If ff is odd, f(−t)=−f(t)f(-t) = -f(t), so ∫−a0f(x)dx=−∫0af(t)dt\int_{-a}^0 f(x) dx = -\int_0^a f(t) dt. Adding this to the second integral yields −∫0af(t)dt+∫0af(x)dx=0-\int_0^a f(t) dt + \int_0^a f(x) dx = 0.

High-Speed ECAT Applications

  • ∫−ππx3cos⁡(2x)dx=0\int_{-\pi}^\pi x^3 \cos(2x) dx = 0 (Because x3x^3 is odd and cos⁡2x\cos 2x is even; odd×even=odd\text{odd} \times \text{even} = \text{odd}).
  • ∫−11x7−3x3+sin⁡xx4+1dx=0\int_{-1}^1 \frac{x^7 - 3x^3 + \sin x}{x^4 + 1} dx = 0 (Every single term in the numerator is odd while the denominator is even).
  • ∫−π/4π/4tan⁡xdx=0\int_{-\pi/4}^{\pi/4} \tan x dx = 0 (Since tan⁡(−x)=−tan⁡x\tan(-x) = -\tan x).

Half-Range Symmetry Property (Integration from 00 to 2a2a)

∫02af(x)dx={2∫0af(x)dxif f(2a−x)=f(x)0if 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}

Example: ∫0πcos⁡5xdx\int_0^\pi \cos^5 x dx. Here 2a=π2a = \pi. Testing: f(π−x)=cos⁡5(π−x)=(−cos⁡x)5=−cos⁡5x=−f(x)f(\pi - x) = \cos^5(\pi - x) = (-\cos x)^5 = -\cos^5 x = -f(x). Because f(π−x)=−f(x)f(\pi - x) = -f(x), the integral is 00 instantly.


5. Periodicity Properties of Definite Integrals

A function f(x)f(x) is periodic with fundamental period T>0T > 0 if f(x+T)=f(x)f(x + T) = f(x) for all xx.

Fundamental Periodic Theorems

  1. Integer Multiple Cycles: ∫0nTf(x)dx=n∫0Tf(x)dx(n∈Z)\int_0^{nT} f(x) dx = n \int_0^T f(x) dx \quad (n \in \mathbb{Z})
  2. Shift Invariance: The integral of a periodic function over a full period is independent of the starting point aa: ∫aa+Tf(x)dx=∫0Tf(x)dx\int_a^{a + T} f(x) dx = \int_0^T f(x) dx
  3. Multiple Shift Cycles: ∫aa+nTf(x)dx=n∫0Tf(x)dx\int_a^{a + nT} f(x) dx = n \int_0^T f(x) dx
  4. Period Shift between Limits: ∫a+nTb+nTf(x)dx=∫abf(x)dx\int_{a + nT}^{b + nT} f(x) dx = \int_a^b f(x) dx

High-Speed ECAT Example: Evaluate ∫020π∣sin⁡x∣dx\int_0^{20\pi} |\sin x| dx. The function ∣sin⁡x∣|\sin x| has fundamental period T=πT = \pi (since ∣sin⁡(x+π)∣=∣−sin⁡x∣=∣sin⁡x∣|\sin(x + \pi)| = |-\sin x| = |\sin x|). Here n=20n = 20. Therefore: ∫020π∣sin⁡x∣dx=20∫0πsin⁡xdx=20[−cos⁡x]0π=20[−(−1)−(−1)]=20(2)=40\int_0^{20\pi} |\sin x| dx = 20 \int_0^\pi \sin x dx = 20 [-\cos x]_0^\pi = 20[-(-1) - (-1)] = 20(2) = 40


6. Leibniz's Rule for Differentiating Under the Integral Sign

On the ECAT, examiners frequently construct limit problems involving definite integrals that yield 00\frac{0}{0} indeterminate forms, requiring L'Hôpital's Rule. Evaluating the derivative of the integral numerator requires Leibniz's Rule.

Formula for Variable Limits of Integration

If f(t)f(t) is continuous and u(x),v(x)u(x), v(x) are differentiable functions of xx:

ddx[∫u(x)v(x)f(t)dt]=f(v(x))⋅v′(x)−f(u(x))⋅u′(x)\frac{d}{dx}\left[ \int_{u(x)}^{v(x)} f(t) dt \right] = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)

Proof: By the Fundamental Theorem of Calculus, ∫u(x)v(x)f(t)dt=F(v(x))−F(u(x))\int_{u(x)}^{v(x)} f(t) dt = F(v(x)) - F(u(x)), where F′(t)=f(t)F'(t) = f(t). Differentiating with respect to xx via the chain rule produces F′(v(x))v′(x)−F′(u(x))u′(x)=f(v(x))v′(x)−f(u(x))u′(x)F'(v(x)) v'(x) - F'(u(x)) u'(x) = f(v(x)) v'(x) - f(u(x)) u'(x).

ECAT Limit Example via Leibniz's Rule

Evaluate lim⁡x→0∫0xsin⁡(t2) dtx3\lim_{x \to 0} \frac{\int_0^{x} \sin(t^2)\, dt}{x^3}.

  • At x=0x = 0, numerator is ∫00=0\int_0^0 = 0 and denominator is 03=00^3 = 0 (indeterminate form 00\frac{0}{0}).
  • Apply L'Hôpital's Rule by differentiating numerator and denominator with respect to xx:
    • ddx[x3]=3x2\frac{d}{dx}[x^3] = 3x^2.
    • By Leibniz's Rule, ddx[∫0xsin⁡(t2) dt]=sin⁡(x2)⋅1−0=sin⁡(x2)\frac{d}{dx}\left[\int_0^{x} \sin(t^2)\, dt\right] = \sin(x^2) \cdot 1 - 0 = \sin(x^2).
  • The limit becomes: lim⁡x→0sin⁡(x2)3x2=13lim⁡u→0sin⁡uu=13\lim_{x \to 0} \frac{\sin(x^2)}{3x^2} = \frac{1}{3} \lim_{u \to 0} \frac{\sin u}{u} = \frac{1}{3}

Trap: If an upper limit such as x2x^2 is substituted into t\sqrt{t}, remember that x2=∣x∣\sqrt{x^2} = |x|, not xx. Ignoring the absolute value can make a two-sided limit look as if it exists when the left-hand and right-hand limits actually differ.

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Decision Flowchart for Definite Integral Evaluation
Test Your Knowledge

What is the exact value of the definite integral ∫0π/2sin⁡xsin⁡x+cos⁡xdx\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx?

A

π4\frac{\pi}{4}

B

π2\frac{\pi}{2}

C

11

D

00

Test Your Knowledge

Let F(x)=∫1x31+t2dtF(x) = \int_1^{x^3} \sqrt{1 + t^2} dt. What is the numerical value of F′(1)F'(1)?

A

2\sqrt{2}

B

222\sqrt{2}

C

00

D

323\sqrt{2}

Test Your Knowledge

What is the value of the definite integral ∫−π/2π/2(x3cos⁡x+sin⁡5x+cos⁡x)dx\int_{-\pi/2}^{\pi/2} (x^3 \cos x + \sin^5 x + \cos x) dx?

A

00

B

22

C

11

D

π\pi

Sections you finish are checked off in the contents.