{"id":388,"date":"2025-02-13T19:44:31","date_gmt":"2025-02-13T19:44:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-fundamental-theorem-of-calculus-learn-it-2\/"},"modified":"2025-02-13T19:44:31","modified_gmt":"2025-02-13T19:44:31","slug":"the-fundamental-theorem-of-calculus-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-fundamental-theorem-of-calculus-learn-it-2\/","title":{"raw":"The Fundamental Theorem of Calculus: Learn It 2","rendered":"The Fundamental Theorem of Calculus: Learn It 2"},"content":{"raw":"\n

Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives<\/h2>\n

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas.<\/p>\n

The theorem is comprised of two parts. The first part, which is stated here, establishes the relationship between differentiation and integration.<\/p>\n

\n

Fundamental Theorem of Calculus, Part 1<\/h3>\n

If [latex]f(x)[\/latex] is continuous over an interval [latex]\\left[a,b\\right],[\/latex] and the function [latex]F(x)[\/latex] is defined by<\/p>\n

[latex]F(x)={\\displaystyle\\int }_{a}^{x}f(t)dt,[\/latex]<\/div>\n

then [latex]{F}^{\\prime }(x)=f(x)[\/latex] over [latex]\\left[a,b\\right].[\/latex]<\/p>\n<\/section>\n

Before we look at the proof, let's clarify a few points:<\/p>\n

    \n\t
  1. \n

    Notation<\/strong>: We define [latex]F(x) [\/latex] as the definite integral of [latex]f(t)[\/latex] from [latex]a[\/latex] to [latex]x[\/latex]. This might seem confusing because we've said a definite integral is a number. However, [latex]F(x)[\/latex] is a function that gives the value of the definite integral for each [latex]x[\/latex].<\/p>\n<\/li>\n\t

  2. Implications<\/strong>: The Fundamental Theorem of Calculus is crucial because it shows that integration and differentiation are inverse processes. It guarantees that any continuous function [latex]f(x)[\/latex] has an antiderivative [latex]F(x)[\/latex].<\/li>\n<\/ol>\n
    \n

    Proof<\/strong><\/p>\n

    \n

    Applying the definition of the derivative, we have<\/p>\n

    [latex]\\begin{array}{}{F}^{\\prime }(x)\\hfill & =\\underset{h\\to 0}{\\text{lim}}\\dfrac{F(x+h)-F(x)}{h}\\hfill \\\\ & =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}\\left[{\\displaystyle\\int }_{a}^{x+h}f(t)dt-{\\displaystyle\\int }_{a}^{x}f(t)dt\\right]\\hfill \\\\ & =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}\\left[{\\displaystyle\\int }_{a}^{x+h}f(t)dt+{\\displaystyle\\int }_{x}^{a}f(t)dt\\right]\\hfill \\\\ & =\\underset{h\\to 0}{\\text{lim}}\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(t)dt.\\hfill \\end{array}[\/latex]<\/center>\n

    Looking carefully at this last expression, we see [latex]\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(t)dt[\/latex] is just the average value of the function [latex]f(x)[\/latex] over the interval [latex]\\left[x,x+h\\right].[\/latex] Therefore, by the mean value theorem for integrals, there is some number [latex]c[\/latex] in [latex]\\left[x,x+h\\right][\/latex] such that<\/p>\n

    [latex]\\dfrac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(x)dx=f(c).[\/latex]<\/center>\n

    In addition, since [latex]c[\/latex] is between [latex]x[\/latex] and [latex]x+h[\/latex], [latex]c[\/latex] approaches [latex]x[\/latex] as [latex]h[\/latex] approaches zero. Also, since [latex]f(x)[\/latex] is continuous, we have [latex]\\underset{h\\to 0}{\\text{lim}}f(c)=\\underset{c\\to x}{\\text{lim}}f(c)=f(x).[\/latex]<\/p>\n

    Putting all these pieces together, we have<\/p>\n

    [latex]\\begin{array}{}{F}^{\\prime }(x)\\hfill & =\\underset{h\\to 0}{\\text{lim}}\\frac{1}{h}{\\displaystyle\\int }_{x}^{x+h}f(x)dx\\hfill \\\\ & =\\underset{h\\to 0}{\\text{lim}}f(c)\\hfill \\\\ & =f(x),\\hfill \\end{array}[\/latex]<\/center>\n

    and the proof is complete.<\/p>\n

    [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

    \n

    Use the first part of the Fundamental Theorem of Calculus to find the derivative of<\/p>\n

    [latex]g(x)={\\displaystyle\\int }_{1}^{x}\\dfrac{1}{{t}^{3}+1}dt.[\/latex]<\/div>\n

    [reveal-answer q=\"fs-id1170572346816\"]Show Solution[\/reveal-answer]
    \n[hidden-answer a=\"fs-id1170572346816\"]<\/p>\n

    According to the Fundamental Theorem of Calculus, the derivative is given by<\/p>\n

    [latex]{g}^{\\prime }(x)=\\frac{1}{{x}^{3}+1}.[\/latex][\/hidden-answer]<\/div>\n<\/section>\n
    \n

    Let [latex]F(x)={\\displaystyle\\int }_{1}^{\\sqrt{x}} \\sin tdt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n

    [reveal-answer q=\"fs-id1170572103655\"]Show Solution[\/reveal-answer]
    \n[hidden-answer a=\"fs-id1170572103655\"]\n\n

    Letting [latex]u(x)=\\sqrt{x},[\/latex] we have [latex]F(x)={\\displaystyle\\int }_{1}^{u(x)} \\sin tdt.[\/latex] Thus, by the Fundamental Theorem of Calculus and the chain rule,<\/p>\n

    [latex]\\begin{array}{}\\\\ {F}^{\\prime }(x)\\hfill & = \\sin (u(x))\\frac{du}{dx}\\hfill \\\\ & = \\sin (u(x))\u00b7(\\frac{1}{2}{x}^{-1\\text{\/}2})\\hfill \\\\ & =\\frac{ \\sin \\sqrt{x}}{2\\sqrt{x}}.\\hfill \\end{array}[\/latex]<\/div>\n
    [\/hidden-answer]<\/div>\n<\/div>\n<\/section>\n
    \n

    Let [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Find [latex]{F}^{\\prime }(x).[\/latex]<\/p>\n

    [reveal-answer q=\"fs-id1170572142330\"]Show Solution[\/reveal-answer]
    \n[hidden-answer a=\"fs-id1170572142330\"]\n\n

    We have [latex]F(x)={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt.[\/latex] Both limits of integration are variable, so we need to split this into two integrals. We get<\/p>\n

    [latex]\\begin{array}{}\\\\ F(x)\\hfill & ={\\displaystyle\\int }_{x}^{2x}{t}^{3}dt\\hfill \\\\ & ={\\displaystyle\\int }_{x}^{0}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\hfill \\\\ & =\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt+{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt.\\hfill \\end{array}[\/latex]<\/div>\n

    Differentiating the first term, we obtain<\/p>\n

    [latex]\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]=\\text{\u2212}{x}^{3}.[\/latex]<\/div>\n

    Differentiating the second term, we first let [latex]u(x)=2x.[\/latex] Then,<\/p>\n

    [latex]\\begin{array}{}\\\\ \\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill & =\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{u(x)}{t}^{3}dt\\right]\\hfill \\\\ & ={(u(x))}^{3}\\frac{du}{dx}\\hfill \\\\ & ={(2x)}^{3}\u00b72\\hfill \\\\ & =16{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n

    Thus,<\/p>\n

    [latex]\\begin{array}{}\\\\ \\\\ {F}^{\\prime }(x)\\hfill & =\\frac{d}{dx}\\left[\\text{\u2212}{\\displaystyle\\int }_{0}^{x}{t}^{3}dt\\right]+\\frac{d}{dx}\\left[{\\displaystyle\\int }_{0}^{2x}{t}^{3}dt\\right]\\hfill \\\\ & =\\text{\u2212}{x}^{3}+16{x}^{3}\\hfill \\\\ & =15{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n

    Watch the following video to see the worked solution to this example.<\/p>\n