Integration BasicsTopic #21 of 32

Fundamental Theorem of Calculus

The connection between differentiation and integration.

Overview

The Fundamental Theorem of Calculus (FTC) connects differentiation and integration, showing they are inverse operations.

Part 1: Differentiation of Integrals

If ff is continuous on [a,b][a, b], then the function:

g(x)=axf(t)dtg(x) = \int_a^x f(t) \, dt

is continuous on [a,b][a, b], differentiable on (a,b)(a, b), and:

g(x)=f(x)g'(x) = f(x)

Or equivalently:

ddx[axf(t)dt]=f(x)\frac{d}{dx} \left[\int_a^x f(t) \, dt\right] = f(x)

With Variable Upper Limit

If the upper limit is a function u(x)u(x):

ddx[au(x)f(t)dt]=f(u(x))u(x)\frac{d}{dx} \left[\int_a^{u(x)} f(t) \, dt\right] = f(u(x)) \cdot u'(x)

With Variable Lower Limit

ddx[v(x)bf(t)dt]=f(v(x))v(x)\frac{d}{dx} \left[\int_{v(x)}^b f(t) \, dt\right] = -f(v(x)) \cdot v'(x)

Both Limits Variable

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

Part 2: Evaluation of Integrals

If ff is continuous on [a,b][a, b] and FF is any antiderivative of ff, then:

abf(x)dx=F(b)F(a)\int_a^b f(x) \, dx = F(b) - F(a)

This is often written as:

abf(x)dx=[F(x)]ab=F(x)ab\int_a^b f(x) \, dx = \left[F(x)\right]_a^b = F(x)\Big|_a^b

The Connection

  • Part 1: Differentiation undoes integration
  • Part 2: Integration undoes differentiation (up to a constant)

Together: axf(t)dt\int_a^x f(t) \, dt is an antiderivative of f(x)f(x)

Examples

Example 1: FTC Part 1 (basic)

ddx[0xcos(t)dt]=cos(x)\frac{d}{dx} \left[\int_0^x \cos(t) \, dt\right] = \cos(x)

Example 2: FTC Part 1 (with chain rule)

ddx[1x2etdt]=ex22x=2xex2\frac{d}{dx} \left[\int_1^{x^2} e^t \, dt\right] = e^{x^2} \cdot 2x = 2xe^{x^2}

Example 3: FTC Part 1 (both limits variable)

ddx[xx3sin(t)dt]=sin(x3)3x2sin(x)1\frac{d}{dx} \left[\int_x^{x^3} \sin(t) \, dt\right] = \sin(x^3) \cdot 3x^2 - \sin(x) \cdot 1

=3x2sin(x3)sin(x)= 3x^2 \sin(x^3) - \sin(x)

Example 4: FTC Part 2 (polynomial)

13(2x1)dx=[x2x]13\int_1^3 (2x - 1) \, dx = \left[x^2 - x\right]_1^3

=(93)(11)=6= (9 - 3) - (1 - 1) = 6

Example 5: FTC Part 2 (exponential)

0ln(2)exdx=[ex]0ln(2)\int_0^{\ln(2)} e^x \, dx = \left[e^x\right]_0^{\ln(2)}

=eln(2)e0=21=1= e^{\ln(2)} - e^0 = 2 - 1 = 1

Example 6: FTC Part 2 (trigonometric)

0π/4sec2(x)dx=[tan(x)]0π/4\int_0^{\pi/4} \sec^2(x) \, dx = \left[\tan(x)\right]_0^{\pi/4}

=tan(π4)tan(0)=10=1= \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1

Example 7: Finding a function

If 0xf(t)dt=x2+2x\int_0^x f(t) \, dt = x^2 + 2x, find f(x)f(x).

By FTC Part 1:

f(x)=ddx[x2+2x]=2x+2f(x) = \frac{d}{dx}[x^2 + 2x] = 2x + 2

Check: 0x(2t+2)dt=[t2+2t]0x=x2+2x\int_0^x (2t + 2) \, dt = \left[t^2 + 2t\right]_0^x = x^2 + 2x

Example 8: Proof that integration is inverse of differentiation

Let F(x)=axf(t)dtF(x) = \int_a^x f(t) \, dt

By FTC Part 1: F(x)=f(x)F'(x) = f(x)

Therefore FF is an antiderivative of ff.

By FTC Part 2:

abf(x)dx=F(b)F(a)=abf(t)dtaaf(t)dt\int_a^b f(x) \, dx = F(b) - F(a) = \int_a^b f(t) \, dt - \int_a^a f(t) \, dt

This confirms the two parts are consistent.

Mean Value Theorem for Integrals

If ff is continuous on [a,b][a, b], there exists cc in [a,b][a, b] such that:

abf(x)dx=f(c)(ba)\int_a^b f(x) \, dx = f(c)(b - a)

This connects to the average value: f(c)=favgf(c) = f_{\text{avg}}