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4.2 Fundamental Theorem of Calculus

Let \(f: [a, b] \rightarrow \mathbb{R}\) be a continuous function and \(G: [a, b] \rightarrow \mathbb{R}\) defined as \(G(x) = \int_{a}^{x} f(t) dt\)
Then \(G\) is differentiable on \((a, b)\) and \(G' = f\) on \([a, b]\).

Let \(f: [a, b] \rightarrow \mathbb{R}\) be integrable and F such that \(F' = f\). Then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\).

  1. Compute \(\frac{dy}{dx}\) for:

    1. \(y = \int_{0}^{x} t^2 dt\)

      \(\frac{dy}{dx} = \frac{d}{dx} \int_{0}^{x} t^2 dt\)

      By FTC: \(=\) \(x^2\)

    2. \(y = \int_{x}^{5} \cos(t^2) dt\)

      \(\frac{dy}{dx} = \frac{d}{dx} \int_{x}^{5} \cos(t^2) dt\) \(= \frac{d}{dx} \left( - \int_{5}^{x} \cos(t^2) dt \right)\) \(=- \frac{d}{dx} \int_{5}^{x} \cos(t^2) dt\). By FTC: \(= -\cos(x^2)\)

    3. \(y = \int_{1}^{x^3} t \sin(2t) dt\)

      The upper limit of integration is not \(x\) but \(x^3\). This makes y a composition of functions.

      \(y = \int_{1}^{u} t \sin(2t) dt\), \(u = x^3\)

      Now to differentiate, we apply the chain rule.

      \(\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\) (chain rule) \(= u \sin(2u) 3x^2\) (FTC) \(= x^3 \sin(2x^3) 3x^2\) \(= 3x^5 \sin(2x^3)\)

    4. \(y = \int_{1}^{x} \sqrt{t} dt - \int_{4}^{x} \sqrt{t} dt\)
      ​\(\frac{d}{dx}(\int_{1}^{x} \sqrt{t} dt - \int_{4}^{x} \sqrt{t} dt )=\) \(\frac{d}{dx}\int_{1}^{x}\sqrt{t}dt - \frac{d}{dx}\int_{4}^{x}\sqrt{t}dt\). By FTC:\(=\)​\(\sqrt{x} - \sqrt{x} = 0\)

      \(y= \int_{1}^{x}\sqrt{t}dt + \int_{x}^{4}\sqrt{t}dt\) (\(c \in [a,b]\)) \(=\int_{1}^{4}\sqrt{t}dt\) \(= \int_{1}^{4}t^{1/2}dt\) \(= \frac{2}{3}t^{3/2} |_{1}^{4} = \frac{2}{3} 4^{3/2} - \frac{2}{3}(1)^{3/2} \in \mathbb{R}\)

    5. \(y = (\int_{0}^{x} \cos t dt )^3\)
      This is a composition. Let \(y = u^3, u = \int_{0}^{x} \cos t dt\)
      By chain rule: \(\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = 3u^2 \cdot \cos x\) \(= 3(\int_{0}^{x} \cos t dt )^2 \cos x\) \(= 3(\sin(x) - \sin(0) )^2 \cos x\)
      ​\(= 3 \sin^2(x) \cos(x)\)

  2. For any continuous function \(f:[a,b] \rightarrow \mathbb{R}\) and any differentiable functions \(g\) and \(h\) that take values in \([a,b]\), we have \(\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x))h'(x) - f(g(x))g'(x).\)
    (In particular if \(g(x) = constant\) and \(h(x) = x\) )

    Define \(y=\int_{a}^{x}f(t)dt\leadsto\frac{dy}{dx}=f(x)\)

    \(\int_{g(x)}^{h(x)}f(t)dt=\int_{a}^{h(x)}f(t)dt-\int_{a}^{g(x)}f(t)dt=y(h(x))-y(g (x))\)

    Now, using chain rule \(\frac{d}{dx}\int_{g(x)}^{h(x)}f(t)dt=\frac{d}{dx}(y(h(x))-y(g(x)))\)
    ​\(=y^{\prime}(h(x))h^{\prime}(x)-y^{\prime}(g(x))g^{\prime}(x)\)
    ​\(= f(h(x))h'(x) - f(g(x))g'(x).\)

  3. Evaluate \(\frac{dy}{dx}\) for:

    1. \(y = \int_{x}^{x^2} \cos(t^2) dt\)

      In this case \(f(t) = \cos(t^2)\), \(g(x) = x\), \(h(x) = x^2\)

      \(\frac{dy}{dx} = f(h(x))h'(x) - f(g(x))g'(x)\) \(= \cos((x^2)^2) 2x - \cos(x^2) \cdot 1\) \(= 2x \cos(x^4) - \cos(x^2)\) 2. \(y = \int_{e^{-x}}^{e^{x}} \sin(\ln(t)) dt\)

      \(f(t) = \sin(\ln(t))\), \(g(x) = e^{-x}\), \(h(x) = e^x\)

      \(\frac{dy}{dx} = \sin(\ln(e^x))e^x - \sin(\ln(e^{-x})) (-e^{-x})\) \(= e^x \sin(x) + e^{-x} \sin(-x)\)

      \(= e^x \sin(x) - e^{-x} \sin(x)\) \(= \sin(x) (e^x - e^{-x})\)

    1. \(\lim_{x \to 0} \frac{1}{x} \int_0^x \sqrt{t^3 + 1} dt\)

      Consider the function \(y(x) = \int_{0}^{x} \sqrt{t^{3} + 1}dt\)

      \(y(0) = 0\) and \(y'(x) = \sqrt{x^{3} + 1}\)

      So \(\lim_{x\to0}\frac{1}{x}\int_{0}^{x}\sqrt{t^{3}+1}dt=\lim_{x\to0}\frac{y(x)}{x}=\lim _{x\to0}\frac{y(x)-y(0)}{x-0}=y^{\prime}(0)=\sqrt{0^{3}+1}=1\)

    2. \(\lim_{x \to a} \frac{1}{e^x - e^a} \int_a^x e^{-t^2} dt\) for any \(a \in \mathbb{R}\)

      Consider \(y(x) = \int_{a}^{x} e^{-t^2}dt\), \(Y(a) = 0\). Then \(y'(x) = e^{-x^2}\)

      \(\lim_{x\to a}\frac{y(x)}{e^{x}-e^{a}}\stackrel{\text{L'Hopital's rule}}{=}\lim_{x\to a}\frac{y^{\prime}(x)}{(e^{x}-e^{a})^{\prime}}=\lim_{x\to a}\frac{e^{-x^2}}{e^{x}} =\lim_{x\to a}e^{-x^2-x}=e^{-a^2-a}\)

Let \(f(x)\) and \(g(x)\) be continuous functions defined on \([a,b]\) where \(f(x) \ge g(x) \ \forall x \in [a,b]\). The area of the region bounded by the curves \(y=f(x)\), \(y=g(x)\) and the lines \(x=a\) and \(x=b\) is \(A = \int_a^b (f(x) - g(x)) dx\).

image​

  1. Determine the area of the region enclosed by

    1. \(y = x^2\) and \(y = \sqrt{x}\)

      image​

      The curves intersect when \(x = 0\) and \(x = 1\).

      \(A = \int_0^1 (\sqrt{x} - x^2) dx = \int_0^1 x^{1/2} dx - \int_0^1 x^2 dx = \frac{2}{3} x^{3/2} \vert_0^1 - \frac{x^3}{3} \vert_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}\) since \(\int x^{a} dx = \frac{x^{a+1}}{a+1}+ c\).

    2. \(y = 2x^2 + 10, y = 4x + 16, x = -2\) and \(x = 5\)

      image​

      \(A = \int_{-2}^{-1} 2x²+10-(4x+16) dx + \int_{-1}^{3} 4x+16-(2x²+10) dx + \int_{3}^{5} 2x²+10-(4x+16) dx\)

      \(= \int_{-2}^{-1} (2x²-4x-6) dx + \int_{-1}^{3} (-2x²+4x+6) dx + \int_{3}^{5} (2x²-4x-6) dy\)

      \(=\frac{2}{3}x^{3}-2x^{2}-6x\left|_{-2}^{-1}+\left(-\frac{2}{3}x^{3}+2x^{2}+6x\right )\right|_{-1}^{3}+\left(\frac{2}{3}x^{3}-2x^{2}-6x\right)\left|_{3}^{5}\right.\)

      \(= \frac{14}{3} + \frac{64}{3} + \frac{64}{3} = \frac{142}{3}\)

    3. \(x = \frac{1}{2}y^2 - 3\) and \(y = x - 1\).

      image

      Method 1:

      \(A = \int_{-3}^{-1} \sqrt{2x+6}-(-\sqrt{2x+6}) dx + \int_{-1}^{5} \sqrt{2x+6}-(x-1) dx\)

      \(= 2 \int_{-3}^{-1} \sqrt{2x+6} dx + \int_{-1}^{5} \sqrt{2x+6}-x+1 dx\)

      \(u = 2x+6\), \(du = 2dx\), \(dx=\frac{du}{2}\)
      If \(x=3\to u=0\). If \(x=1\to u=4\). If \(x=5\to u=16\)

      \(=2\int_{0}^{4}\frac{\sqrt{u}}{2}du+\int_{4}^{16}\frac{\sqrt{u}}{2}du+\int_{-1}^{5} -x+1dx\), check = \(18\)

      Method 2:

      image​

      Calculate in terms of \(y\).
      \(x=\frac{1}{2}y²-3\), \(x=y+1\)

      intersection \(\frac{1}{2}y²-3=y+1 \implies y=4, y=-2\)

      \(A = \int_{-2}^{4} y+1-(\frac{1}{2}y²-3) dy = \int_{-2}^{4} y+1-\frac{1}{2}y²+3 dy\) \(= \frac{y²}{2}-\frac{1}{6}y³+4y|_{-2}^{4}\), check = \(18\)