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Home/Calculus/Chapter 6/Fundamental Theorem of Calculus
CALC-6.3
4-5 hours

Fundamental Theorem of Calculus

The most important theorem in calculus: the Newton-Leibniz formula reveals that differentiation and integration are inverse operations.

Learning Objectives
Define and analyze variable upper limit integrals
Prove continuity of the integral function F(x)
Establish differentiability: F'(x) = f(x)
Prove the Newton-Leibniz formula
Apply FTC to compute definite integrals
Differentiate integrals with variable limits
Evaluate limits using L'Hôpital with integrals

1. Variable Upper Limit Integral

Definition 6.8: Variable Upper Limit Integral

If fR[a,b]f \in R[a, b], define the integral function:

F(x)=axf(t)dt,x[a,b]F(x) = \int_a^x f(t)\,dt, \quad x \in [a, b]

Note: We use a different variable tt inside to avoid confusion with the upper limit xx.

Theorem 6.16: Continuity of the Integral Function

If fR[a,b]f \in R[a, b], then F(x)=axf(t)dtF(x) = \int_a^x f(t)\,dt is continuous on [a,b][a, b].

In fact, FF is Lipschitz continuous: F(x)F(y)Mxy|F(x) - F(y)| \leq M|x - y| where M=supfM = \sup|f|.

Proof of Theorem 6.16:

For x,y[a,b]x, y \in [a, b] with x<yx < y:

F(y)F(x)=xyf(t)dtxyf(t)dtM(yx)|F(y) - F(x)| = \left| \int_x^y f(t)\,dt \right| \leq \int_x^y |f(t)|\,dt \leq M(y - x)

This shows FF is Lipschitz continuous, hence uniformly continuous.

2. FTC Part 1: Derivative of the Integral

Theorem 6.17: Fundamental Theorem of Calculus (Part 1)

Let fR[a,b]f \in R[a, b] and F(x)=axf(t)dtF(x) = \int_a^x f(t)\,dt.

If ff is continuous at x0[a,b]x_0 \in [a, b], then FF is differentiable at x0x_0 and:

F(x0)=f(x0)F'(x_0) = f(x_0)

In short: “The derivative of an integral is the integrand.”

Proof of Theorem 6.17:

Consider the difference quotient for h>0h > 0:

F(x0+h)F(x0)h=1hx0x0+hf(t)dt\frac{F(x_0 + h) - F(x_0)}{h} = \frac{1}{h} \int_{x_0}^{x_0+h} f(t)\,dt

Since ff is continuous at x0x_0, for any ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that:

tx0<δf(t)f(x0)<ϵ|t - x_0| < \delta \Rightarrow |f(t) - f(x_0)| < \epsilon

For 0<h<δ0 < h < \delta:

F(x0+h)F(x0)hf(x0)=1hx0x0+h[f(t)f(x0)]dt\left| \frac{F(x_0 + h) - F(x_0)}{h} - f(x_0) \right| = \left| \frac{1}{h} \int_{x_0}^{x_0+h} [f(t) - f(x_0)]\,dt \right|
1hx0x0+hf(t)f(x0)dt<1hϵh=ϵ\leq \frac{1}{h} \int_{x_0}^{x_0+h} |f(t) - f(x_0)|\,dt < \frac{1}{h} \cdot \epsilon \cdot h = \epsilon

Similarly for h<0h < 0. Thus F(x0)=f(x0)F'(x_0) = f(x_0).

Corollary 6.4: Continuous Functions Have Antiderivatives

If fC[a,b]f \in C[a, b], then F(x)=axf(t)dtF(x) = \int_a^x f(t)\,dt is an antiderivative of ff on [a,b][a, b].

Remark 6.5: Antiderivatives vs. Integrability

Important distinctions:

  • An integrable function may NOT have an elementary antiderivative (e.g., ex2e^{-x^2})
  • A function with an antiderivative may NOT be integrable (pathological examples exist)
  • For continuous functions, both directions work!

3. Newton-Leibniz Formula (FTC Part 2)

Theorem 6.18: Newton-Leibniz Formula

Let fR[a,b]f \in R[a, b] and suppose FF is an antiderivative of ff on [a,b][a, b] (i.e., F(x)=f(x)F'(x) = f(x)). Then:

abf(x)dx=F(b)F(a)=:F(x)ab\int_a^b f(x)\,dx = F(b) - F(a) =: F(x)\Big|_a^b
Proof of Theorem 6.18:

Take any partition Δ:a=x0<x1<<xn=b\Delta: a = x_0 < x_1 < \cdots < x_n = b.

By the Mean Value Theorem, for each kk there exists ξk(xk1,xk)\xi_k \in (x_{k-1}, x_k) such that:

F(xk)F(xk1)=F(ξk)(xkxk1)=f(ξk)ΔxkF(x_k) - F(x_{k-1}) = F'(\xi_k)(x_k - x_{k-1}) = f(\xi_k)\Delta x_k

Summing over all kk (telescoping sum):

F(b)F(a)=k=1n[F(xk)F(xk1)]=k=1nf(ξk)ΔxkF(b) - F(a) = \sum_{k=1}^n [F(x_k) - F(x_{k-1})] = \sum_{k=1}^n f(\xi_k)\Delta x_k

Taking Δ0\|\Delta\| \to 0, the right side converges to abf(x)dx\int_a^b f(x)\,dx.

Remark 6.6: Evaluation Notation

The notation F(x)abF(x)\big|_a^b or [F(x)]ab[F(x)]_a^b means:

F(x)ab=F(b)F(a)F(x)\Big|_a^b = F(b) - F(a)

This is read as “F(x)F(x) evaluated from aa to bb”.

Example 6.12: Basic Application of Newton-Leibniz

Compute: 01x2dx\int_0^1 x^2\,dx

Antiderivative: F(x)=x33F(x) = \frac{x^3}{3}

01x2dx=x3301=130=13\int_0^1 x^2\,dx = \frac{x^3}{3}\Big|_0^1 = \frac{1}{3} - 0 = \frac{1}{3}

4. Differentiation with Variable Limits

Corollary 6.5: General Leibniz Rule

If ϕ(x)\phi(x) and ψ(x)\psi(x) are differentiable and ff is continuous:

ddxϕ(x)ψ(x)f(t)dt=f(ψ(x))ψ(x)f(ϕ(x))ϕ(x)\frac{d}{dx} \int_{\phi(x)}^{\psi(x)} f(t)\,dt = f(\psi(x))\psi'(x) - f(\phi(x))\phi'(x)
Proof of Corollary 6.5:

Let G(u)=auf(t)dtG(u) = \int_a^u f(t)\,dt. Then:

ϕ(x)ψ(x)f(t)dt=G(ψ(x))G(ϕ(x))\int_{\phi(x)}^{\psi(x)} f(t)\,dt = G(\psi(x)) - G(\phi(x))

By the chain rule:

ddx[G(ψ(x))G(ϕ(x))]=G(ψ(x))ψ(x)G(ϕ(x))ϕ(x)\frac{d}{dx}[G(\psi(x)) - G(\phi(x))] = G'(\psi(x))\psi'(x) - G'(\phi(x))\phi'(x)

Since G(u)=f(u)G'(u) = f(u), the result follows.

Example 6.13: Variable Limits

Find: ddxxx2et2dt\frac{d}{dx} \int_x^{x^2} e^{-t^2}\,dt

Here ϕ(x)=x\phi(x) = x, ψ(x)=x2\psi(x) = x^2, f(t)=et2f(t) = e^{-t^2}.

ddxxx2et2dt=ex42xex21=2xex4ex2\frac{d}{dx} \int_x^{x^2} e^{-t^2}\,dt = e^{-x^4} \cdot 2x - e^{-x^2} \cdot 1 = 2xe^{-x^4} - e^{-x^2}
Example 6.14: Limit Evaluation

Evaluate: limx00xcost2dtx\lim_{x \to 0} \frac{\int_0^x \cos t^2\,dt}{x}

This is 00\frac{0}{0} form. By L'Hôpital's Rule:

limx00xcost2dtx=limx0cosx21=cos0=1\lim_{x \to 0} \frac{\int_0^x \cos t^2\,dt}{x} = \lim_{x \to 0} \frac{\cos x^2}{1} = \cos 0 = 1

5. Applications and Examples

Example 6.15: Converting Sums to Integrals

Evaluate: limnn(1n2+1+1n2+4++1n2+n2)\lim_{n \to \infty} n\left( \frac{1}{n^2+1} + \frac{1}{n^2+4} + \cdots + \frac{1}{n^2+n^2} \right)

=limnk=1nnn2+k2=limnk=1n11+(k/n)21n= \lim_{n \to \infty} \sum_{k=1}^n \frac{n}{n^2 + k^2} = \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{1 + (k/n)^2} \cdot \frac{1}{n}

This is a Riemann sum for f(x)=11+x2f(x) = \frac{1}{1+x^2} on [0,1][0, 1]:

=01dx1+x2=arctanx01=π4= \int_0^1 \frac{dx}{1+x^2} = \arctan x\Big|_0^1 = \frac{\pi}{4}
Example 6.16: Uniform Convergence Application

Problem: If fC1[0,1]f \in C^1[0, 1], prove:

limnk=1n[f(x+kn2+k2)f(x)]=f(x)ln22\lim_{n \to \infty} \sum_{k=1}^n \left[ f\left( x + \frac{k}{n^2+k^2} \right) - f(x) \right] = f'(x) \cdot \frac{\ln 2}{2}

Sketch:

Use Mean Value Theorem and recognize the Riemann sum structure for 01dt1+t2=π4\int_0^1 \frac{dt}{1+t^2} = \frac{\pi}{4}... (detailed proof involves careful analysis).

6. Common Mistakes

❌ Wrong: ddxabf(t)dt=f(x)\frac{d}{dx}\int_a^b f(t)\,dt = f(x)

The integral from aa to bb (both constants) is a constant. Its derivative is 0!

❌ Forgetting the Chain Rule

ddxax2f=f(x2)2x\frac{d}{dx}\int_a^{x^2} f = f(x^2) \cdot 2x, not just f(x2)f(x^2)!

❌ Wrong sign for lower limit

ddxxbf=f(x)\frac{d}{dx}\int_x^b f = -f(x) (note the minus sign!)

❌ Assuming all integrals have elementary antiderivatives

ex2dx\int e^{-x^2}dx has no elementary form. The integral exists but can't be expressed simply.

7. Additional Examples

Example 6.17: Computing Definite Integrals

Evaluate: 0π/2cosxdx\int_0^{\pi/2} \cos x\,dx

Solution:

Antiderivative of cosx\cos x is sinx\sin x.

0π/2cosxdx=sinx0π/2=sinπ2sin0=10=1\int_0^{\pi/2} \cos x\,dx = \sin x\Big|_0^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1
Example 6.18: Polynomial Integration

Evaluate: 12(3x22x+1)dx\int_1^2 (3x^2 - 2x + 1)\,dx

Solution:

Antiderivative: F(x)=x3x2+xF(x) = x^3 - x^2 + x

12(3x22x+1)dx=[x3x2+x]12\int_1^2 (3x^2 - 2x + 1)\,dx = [x^3 - x^2 + x]_1^2
=(84+2)(11+1)=61=5= (8 - 4 + 2) - (1 - 1 + 1) = 6 - 1 = 5
Example 6.19: Exponential and Logarithmic

Evaluate: 1e1xdx\int_1^e \frac{1}{x}\,dx and 01exdx\int_0^1 e^x\,dx

Solutions:

1e1xdx=lnx1e=lneln1=10=1\int_1^e \frac{1}{x}\,dx = \ln x\Big|_1^e = \ln e - \ln 1 = 1 - 0 = 1
01exdx=ex01=e1\int_0^1 e^x\,dx = e^x\Big|_0^1 = e - 1
Example 6.20: Differentiation Under Integral Sign

Find: F(x)F'(x) where F(x)=0x3sin(t2)dtF(x) = \int_0^{x^3} \sin(t^2)\,dt

Solution:

Using the Leibniz rule with ψ(x)=x3\psi(x) = x^3:

F(x)=sin((x3)2)ddx(x3)=sin(x6)3x2=3x2sin(x6)F'(x) = \sin((x^3)^2) \cdot \frac{d}{dx}(x^3) = \sin(x^6) \cdot 3x^2 = 3x^2 \sin(x^6)
Example 6.21: Both Limits Variable

Find: ddxsinxcosxet2dt\frac{d}{dx} \int_{\sin x}^{\cos x} e^{t^2}\,dt

Solution:

Using the Leibniz rule:

=ecos2x(sinx)esin2xcosx= e^{\cos^2 x} \cdot (-\sin x) - e^{\sin^2 x} \cdot \cos x
=sinxecos2xcosxesin2x= -\sin x \cdot e^{\cos^2 x} - \cos x \cdot e^{\sin^2 x}
Example 6.22: L'Hôpital with Integrals

Evaluate: limx00xsint2dtx3\lim_{x \to 0} \frac{\int_0^x \sin t^2\,dt}{x^3}

Solution:

This is 00\frac{0}{0} form. Apply L'Hôpital:

limx0sinx23x2\lim_{x \to 0} \frac{\sin x^2}{3x^2}

Still 00\frac{0}{0}. Apply again or use Taylor: sinx2x2\sin x^2 \approx x^2 for small xx:

=limx0x23x2=13= \lim_{x \to 0} \frac{x^2}{3x^2} = \frac{1}{3}
Example 6.23: Finding Functions from Integral Equations

Problem: Find f(x)f(x) if 0xf(t)dt=x2+xsinx\int_0^x f(t)\,dt = x^2 + x\sin x

Solution:

By FTC Part 1, differentiate both sides:

f(x)=ddx(x2+xsinx)=2x+sinx+xcosxf(x) = \frac{d}{dx}(x^2 + x\sin x) = 2x + \sin x + x\cos x

Verify: 00f(t)dt=0\int_0^0 f(t)\,dt = 0 and RHS at x=0x=0 is 0+0=00 + 0 = 0. ✓

8. Practice Problems

Problem Set A: Newton-Leibniz Formula

  1. 1.Evaluate 0πsinxdx\int_0^{\pi} \sin x\,dx
  2. 2.Evaluate 14xdx\int_1^4 \sqrt{x}\,dx
  3. 3.Evaluate 0111+x2dx\int_0^1 \frac{1}{1+x^2}\,dx
  4. 4.Evaluate 11xdx\int_{-1}^1 |x|\,dx

Problem Set B: Differentiation of Integrals

  1. 5.Find ddx1xlntdt\frac{d}{dx}\int_1^x \ln t\,dt
  2. 6.Find ddx0x2et2dt\frac{d}{dx}\int_0^{x^2} e^{-t^2}\,dt
  3. 7.Find ddxx2x1tdt\frac{d}{dx}\int_x^{2x} \frac{1}{t}\,dt
  4. 8.Find ddxxx2cos(t3)dt\frac{d}{dx}\int_{\sqrt{x}}^{x^2} \cos(t^3)\,dt

Problem Set C: Limits and Applications

  1. 9.Evaluate limx00xet2dtx\lim_{x \to 0} \frac{\int_0^x e^{t^2}\,dt}{x}
  2. 10.Evaluate limx00xtantdtx2\lim_{x \to 0} \frac{\int_0^x \tan t\,dt}{x^2}
  3. 11.If F(x)=0x(xt)f(t)dtF(x) = \int_0^x (x-t)f(t)\,dt, find F(x)F''(x)
  4. 12.Solve: 0xf(t)dt=x+x1tf(t)dt\int_0^x f(t)\,dt = x + \int_x^1 tf(t)\,dt for ff
Remark 6.7: Answers to Selected Problems

Problem 1: 22

Problem 2: 143\frac{14}{3}

Problem 3: π4\frac{\pi}{4}

Problem 5: lnx\ln x

Problem 6: 2xex42xe^{-x^4}

Problem 9: 11

9. Deeper Insights

Why FTC is Revolutionary

Unification

Before FTC, differentiation (tangent problem) and integration (area problem) were seen as separate. FTC reveals they are inverse operations—a profound insight.

Computational Power

Without FTC, computing 01x100dx\int_0^1 x^{100}\,dx would require summing 100 terms and taking limits. With FTC: x10110101=1101\frac{x^{101}}{101}\big|_0^1 = \frac{1}{101}!

Existence of Antiderivatives

FTC Part 1 guarantees that every continuous function has an antiderivative—even if we can't write it in elementary form. The integral F(x)=axfF(x) = \int_a^x f always works!

Physical Interpretation

If v(t)v(t) is velocity, then abv(t)dt=s(b)s(a)\int_a^b v(t)\,dt = s(b) - s(a) is displacement. The rate of change (v=sv = s') and cumulative change are intimately linked.

Theorem 6.19: Generalized FTC

If FF is absolutely continuous on [a,b][a, b], then FF' exists almost everywhere, FL1[a,b]F' \in L^1[a,b], and:

F(x)F(a)=axF(t)dtF(x) - F(a) = \int_a^x F'(t)\,dt

This is the Lebesgue version of FTC, valid for a broader class of functions.

Remark 6.8: Non-elementary Antiderivatives

Some important functions have no elementary antiderivative:

Error function

erf(x)=2π0xet2dt\text{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt

Sine integral

Si(x)=0xsinttdt\text{Si}(x) = \int_0^x \frac{\sin t}{t}\,dt

Logarithmic integral

li(x)=0xdtlnt\text{li}(x) = \int_0^x \frac{dt}{\ln t}

Fresnel integral

S(x)=0xsin(t2)dtS(x) = \int_0^x \sin(t^2)\,dt

10. Historical Context

Isaac Newton (1642-1727)

Developed his “method of fluxions” in the 1660s, recognizing that differentiation and integration are inverse processes. He kept his work largely unpublished until later.

Gottfried Wilhelm Leibniz (1646-1716)

Independently developed calculus in the 1670s-80s. His notation (\int, dxdx) and systematic approach proved more influential. The “Newton-Leibniz formula” honors both.

James Gregory (1638-1675)

Scottish mathematician who stated a geometric version of FTC before Newton and Leibniz, though his work was less complete and systematic.

Isaac Barrow (1630-1677)

Newton's teacher at Cambridge who proved a geometric form of FTC in his “Lectiones Geometricae” (1670), directly influencing Newton's work.

11. Summary of Key Results

ResultFormulaConditions
FTC Part 1ddxaxf(t)dt=f(x)\frac{d}{dx}\int_a^x f(t)\,dt = f(x)ff continuous at xx
Newton-Leibnizabfdx=F(b)F(a)\int_a^b f\,dx = F(b) - F(a)F=fF' = f, fR[a,b]f \in R[a,b]
Leibniz Ruleddxϕψf=f(ψ)ψf(ϕ)ϕ\frac{d}{dx}\int_\phi^\psi f = f(\psi)\psi' - f(\phi)\phi'f,ϕ,ψf, \phi, \psi differentiable
Continuity of FFF(x)F(y)Mxy|F(x) - F(y)| \leq M|x-y|fR[a,b]f \in R[a,b], M=supfM = \sup|f|

12. What's Next?

In the next section, we learn Computation Techniques for definite integrals:

Substitution method for definite integrals
Integration by parts formula
Wallis integrals and reduction formulas
Symmetry tricks for efficient computation

13. Study Tips

Memorize the Two Parts

FTC Part 1: Differentiation undoes integration. FTC Part 2 (Newton-Leibniz): How to evaluate definite integrals using antiderivatives.

Master the Leibniz Rule

For variable limits: differentiate the upper limit (multiply by its derivative), subtract the lower limit term.ddxϕ(x)ψ(x)f(t)dt=f(ψ)ψf(ϕ)ϕ\frac{d}{dx}\int_{\phi(x)}^{\psi(x)} f(t)\,dt = f(\psi)\psi' - f(\phi)\phi'

Check Continuity

FTC Part 1 requires ff to be continuous at the point of differentiation. FTC Part 2 requires an antiderivative to exist.

Practice L'Hôpital Applications

Many limit problems involve 00\frac{0}{0} forms with integrals. Use FTC to differentiate the numerator.

14. Challenge Problems

Challenge 1: Nested Integrals

Find F(x)F''(x) where:

F(x)=0x(0tf(s)ds)dtF(x) = \int_0^x \left(\int_0^t f(s)\,ds\right) dt

Answer: F(x)=f(x)F''(x) = f(x)

Challenge 2: Integral Equation

Solve for ff:

f(x)=1+0xf(t)dtf(x) = 1 + \int_0^x f(t)\,dt

Hint: Differentiate both sides and solve the ODE.

Challenge 3: Limit with Parameter

Evaluate:

limx0+0xsin(t2)dtx3\lim_{x \to 0^+} \frac{\int_0^x \sin(t^2)\,dt}{x^3}

Hint: Apply L'Hôpital twice and use Taylor expansion.

Challenge 4: Generalized Leibniz

For parameter-dependent integrands:

ddxabf(x,t)dt=abfx(x,t)dt\frac{d}{dx}\int_a^b f(x, t)\,dt = \int_a^b \frac{\partial f}{\partial x}(x, t)\,dt

When does this hold? What conditions are needed?

15. Connections to Other Topics

Differential Equations

FTC is key to solving ODEs. y=f(x)y' = f(x) has solution y=f(x)dx+Cy = \int f(x)\,dx + C. Many techniques rely on the inverse relationship between derivatives and integrals.

Physics Applications

Position from velocity: s(t)=s0+0tv(τ)dτs(t) = s_0 + \int_0^t v(\tau)\,d\tau. Work from force: W=abF(x)dxW = \int_a^b F(x)\,dx. FTC unifies these computations.

Probability Theory

CDF from PDF: F(x)=xf(t)dtF(x) = \int_{-\infty}^x f(t)\,dt. By FTC: F(x)=f(x)F'(x) = f(x)—the PDF is the derivative of the CDF.

Complex Analysis

FTC generalizes to complex contour integrals. If F=fF' = f on a region, thenγfdz=F(end)F(start)\int_\gamma f\,dz = F(\text{end}) - F(\text{start}).

16. Quick Reference Card

Key Formulas

FTC 1ddxaxf=f(x)\frac{d}{dx}\int_a^x f = f(x)
FTC 2abf=F(b)F(a)\int_a^b f = F(b) - F(a)
Leibnizf(ψ)ψf(ϕ)ϕf(\psi)\psi' - f(\phi)\phi'

Common Patterns

Upper limit g(x)g(x):

ddxag(x)f(t)dt=f(g(x))g(x)\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)

Lower limit h(x)h(x):

ddxh(x)bf(t)dt=f(h(x))h(x)\frac{d}{dx}\int_{h(x)}^b f(t)\,dt = -f(h(x)) \cdot h'(x)

Properties of F(x)F(x)

  • Continuous if fR[a,b]f \in R[a,b]
  • Differentiable where ff is continuous
  • F(a)=0F(a) = 0

Conditions

  • FTC 1: ff continuous at xx
  • FTC 2: F=fF' = f on [a,b][a,b]
  • Both: fR[a,b]f \in R[a,b]

Common Errors

  • Forgetting chain rule
  • Wrong sign for lower limit
  • Assuming all ff have elementary FF

17. Extended Examples

Example 6.24: Convolution-type Integral

Problem: Find F(x)F'(x) where F(x)=0x(xt)2f(t)dtF(x) = \int_0^x (x - t)^2 f(t)\,dt

Solution:

Expand: (xt)2=x22xt+t2(x-t)^2 = x^2 - 2xt + t^2

F(x)=x20xf(t)dt2x0xtf(t)dt+0xt2f(t)dtF(x) = x^2\int_0^x f(t)\,dt - 2x\int_0^x tf(t)\,dt + \int_0^x t^2 f(t)\,dt

Now differentiate using product rule + FTC:

F(x)=2x0xf+x2f(x)20xtfdt2xxf(x)+x2f(x)F'(x) = 2x\int_0^x f + x^2 f(x) - 2\int_0^x tf\,dt - 2x \cdot xf(x) + x^2f(x)
=2x0xf(t)dt20xtf(t)dt= 2x\int_0^x f(t)\,dt - 2\int_0^x tf(t)\,dt
Example 6.25: Asymptotic Behavior

Problem: Find limx0xet2dtex22x\lim_{x \to \infty} \frac{\int_0^x e^{-t^2}\,dt}{\frac{e^{-x^2}}{2x}}

Solution:

This is 0\frac{\infty}{0}. Rewrite and apply L'Hôpital:

limx2x0xet2dtex2=lim2+2xex22xex2\lim_{x \to \infty} \frac{2x\int_0^x e^{-t^2}\,dt}{e^{-x^2}} = \lim \frac{2\int + 2x \cdot e^{-x^2}}{-2xe^{-x^2}}

The integral grows like π/2\sqrt{\pi}/2, so this limit is 1-1.

Example 6.26: Area Under Parametric Curve

Problem: Find the area under x=t2x = t^2, y=t3y = t^3 for t[0,1]t \in [0, 1]

Solution:

Use A=ydx=01t32tdtA = \int y\,dx = \int_0^1 t^3 \cdot 2t\,dt:

A=012t4dt=2t5501=25A = \int_0^1 2t^4\,dt = \frac{2t^5}{5}\Big|_0^1 = \frac{2}{5}
Fundamental Theorem Quiz
8
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1
The function F(x)=axf(t)dtF(x) = \int_a^x f(t)\,dt is called:
Easy
Not attempted
2
If fR[a,b]f \in R[a,b], then F(x)=axf(t)dtF(x) = \int_a^x f(t)\,dt is:
Easy
Not attempted
3
If ff is continuous at x0x_0, then F(x0)F'(x_0) equals:
Easy
Not attempted
4
The Newton-Leibniz formula states:
Easy
Not attempted
5
ddxax2sintdt\frac{d}{dx}\int_a^{x^2} \sin t\,dt equals:
Medium
Not attempted
6
ddxxx2f(t)dt\frac{d}{dx}\int_{x}^{x^2} f(t)\,dt equals:
Hard
Not attempted
7
limx00xcost2dtx\lim_{x \to 0} \frac{\int_0^x \cos t^2\,dt}{x} equals:
Medium
Not attempted
8
If fR[a,b]f \in R[a,b] but NOT continuous, then F(x)=axfF(x) = \int_a^x f is:
Hard
Not attempted

Frequently Asked Questions

Why is FTC so important?

FTC connects the two main operations of calculus: differentiation and integration are inverse operations. This lets us evaluate definite integrals using antiderivatives instead of limits of Riemann sums.

Does every integrable function have an antiderivative?

No! A function can be integrable without having an antiderivative expressible in elementary functions. Example: e^(-x²) is integrable but has no elementary antiderivative.

What if f is discontinuous?

If f ∈ R[a,b], F(x) = ∫ₐˣ f is still continuous. However, F is only differentiable at points where f is continuous, and F'(x₀) = f(x₀) only at such points.

Can the lower limit also vary?

Yes! Use ∫ₐˣ = -∫ˣₐ. For ∫_{g(x)}^{h(x)} f(t)dt, differentiate as: f(h(x))h'(x) - f(g(x))g'(x).

Why does the notation F(x)|ₐᵇ mean F(b) - F(a)?

This notation (evaluation bar) is shorthand. We 'evaluate at b' then 'subtract evaluation at a'. It comes from the Newton-Leibniz formula.

Key Takeaways

FTC Part 1

ddxaxf(t)dt=f(x)\frac{d}{dx}\int_a^x f(t)\,dt = f(x)

Newton-Leibniz

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

Leibniz Rule

ddxϕψf=f(ψ)ψf(ϕ)ϕ\frac{d}{dx}\int_{\phi}^{\psi} f = f(\psi)\psi' - f(\phi)\phi'

Key Insight

Differentiation and integration are inverse operations!

Continuity of FF

F(x)=axfF(x) = \int_a^x f is always continuous

Existence

Every continuous ff has an antiderivative via FTC

Decision Guide: Which Formula?

Evaluating: Newton-Leibniz with antiderivative
Differentiating: FTC Part 1 + chain rule
Variable limits: Leibniz rule
Limit problems: L'Hôpital + FTC
Integral equations: Differentiate both sides
No elementary form: Define FF via integral

Quick Decision Guide

When to Use FTC Part 1

  • Differentiating integral with variable upper limit
  • Solving for functions from integral equations
  • Proving existence of antiderivatives

When to Use Newton-Leibniz

  • Computing definite integrals with known antiderivative
  • Evaluating areas, volumes, work
  • Finding net change from rate of change

18. Common Exam Patterns

Pattern 1: Direct Differentiation

Given F(x)=ag(x)f(t)dtF(x) = \int_a^{g(x)} f(t)\,dt, find F(x)F'(x).

Solution: Apply chain rule: F(x)=f(g(x))g(x)F'(x) = f(g(x)) \cdot g'(x)

Pattern 2: Limit Problems

Evaluate limxaaxf(t)dtg(x)\lim_{x \to a} \frac{\int_a^x f(t)\,dt}{g(x)} where g(a)=0g(a) = 0.

Solution: L'Hôpital's rule gives f(x)g(x)\frac{f(x)}{g'(x)} evaluated at x=ax = a.

Pattern 3: Integral Equations

Find ff if 0xf(t)dt=G(x)\int_0^x f(t)\,dt = G(x).

Solution: Differentiate both sides: f(x)=G(x)f(x) = G'(x).

Pattern 4: Both Limits Variable

Find ddxϕ(x)ψ(x)f(t)dt\frac{d}{dx}\int_{\phi(x)}^{\psi(x)} f(t)\,dt.

Solution: f(ψ)ψf(ϕ)ϕf(\psi)\psi' - f(\phi)\phi'.

19. Final Remarks

The Fundamental Theorem of Calculus is arguably the most important result in elementary calculus. It unifies two seemingly disparate problems—finding tangent lines (differentiation) and computing areas (integration)—revealing them as inverse operations.

This connection has profound implications throughout mathematics, physics, engineering, and beyond. From computing work and energy to modeling population growth, from signal processing to probability theory, FTC provides the foundational bridge between rates and accumulations.

As you continue your study of calculus, you will see FTC appear again and again—in integration techniques, in differential equations, in multivariable calculus (as the various generalized Stokes' theorems), and in real and complex analysis. Mastering FTC now will pay dividends throughout your mathematical journey.

Key Points to Remember

  • • FTC Part 1: Differentiation undoes integration
  • • FTC Part 2: Integration undoes differentiation (up to constant)
  • • Variable limits require chain rule (Leibniz rule)
  • • Not all integrands have elementary antiderivatives
  • • FTC connects local (derivative) with global (integral)
  • • Continuity of integrand ensures differentiability of integral function
  • • Practice with variable limit problems builds essential skills
  • • The error function erf(x) and Fresnel integrals are defined via FTC