Fundamental Theorem of Calculus

In this wiki, we will see how the two main branches of calculus, differential and integral calculus, are related to each other. While the two might seem to be unrelated to each other, as one arose from the tangent problem and the other arose from the area problem, we will see that the fundamental theorem of calculus does indeed create a link between the two.

We have learned about indefinite integrals, which was the process of finding the antiderivative of a function. In contrast to the indefinite integral, the result of a definite integral will be a number, instead of a function. The definite integral of a function is the signed area under the graph of the function, and is expressed in the form of $\displaystyle{\int_a^b f(x)\, dx}:$

Now, suppose that we formed an area function $S(x)$ in such a way that it is dependent on the function $f(x)$ as

$$S(x)=\int _{ a }^{ x }{ f(t)\, dt },$$

where $f$ is continuous on the interval $[a,b]$. Now, suppose we wanted to find the the rate of change of the area with respect to $x$:

We can see from the figure above that the area of the shaded region is equal to the area under the curve $f(t)$ from $a$ to $x+\Delta x$ minus the area under $f(t)$ from $a$ to $x$. Thus,

$$\begin{aligned} \Delta S&=A(x+\Delta x)-A(x)\\\\ \frac{\Delta S}{\Delta x}&=\frac{A(x+\Delta x)-A(x)}{\Delta x}. \end{aligned}$$

So, the rate of change of area becomes

$$S'(x)=\frac{dS}{dx}=\lim_ {\Delta x\rightarrow 0 } \frac { S(x+\Delta x)-S(x) }{ \Delta x } .$$

We know that there is an $\overline{x}$ found between $x$ and $x+\Delta x$ such that the area of the shaded region is equal to $f(\overline{x})\Delta x$:

$$\begin{aligned} S'(x) &=\lim_{\Delta x\rightarrow 0 }\frac { S(x+\Delta x)-S(x) }{ \Delta x } \\ &=\lim_{\Delta x\rightarrow 0 }\frac { f(\overline { x } )\Delta x }{ \Delta x } \\ &=\lim_{\Delta x\rightarrow 0 } f(\overline { x } )\\ &=f(x). \end{aligned}$$

The last step is true because, as $\Delta x\rightarrow 0$, anything found between $x$ and $x+\Delta x$ approaches $x$. So, now we are ready to state the first fundamental theorem of calculus:

If $f$ is continuous on $[a,b]$, then the function defined by

$$S(x)=\int _{ a }^{ x }{ f(t)\, dt }$$

is continuous on $[a,b]$ and differentiable on $(a,b)$, and $S'(x)=f(x)$.

So basically integration is the opposite of differentiation. More clearly, the first fundamental theorem of calculus can be rewritten in Leibniz notation as

$$\frac { d }{ dx } \int _{ a }^{ x }{ f(t)\, dt=f(x) }.$$

Find the derivative of $\displaystyle k(x)=\int _{ 2 }^{ x }{ ({ 4}^{ t }+t)\, dt }$.

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The function $f$ is continuous, so from the first fundamental theorem of calculus we have

$$k'(x)={ 4 }^{ x }+x. \ _\square$$

What is the derivative of $\displaystyle h(x)=\int_{ 2 }^{ { x }^{ 2 } }{ \frac {1}{ 1+{ t }^{ 2 } }\, dt }?$

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We use the first fundamental theorem of calculus in accordance with the chain-rule to solve this.

Let $u=x^{2}$, then

$$\begin{aligned} \frac{d}{dx} \int_{2}^{x^2}{\frac{1}{1+t^2}\, dt} &= \frac { d }{ du } \left[ \int _{ 1 }^{ u }{ \frac { 1 }{ 1+{ t }^{ 2 } } dt} \right] \cdot \frac { du }{ dx } \\ &=\frac { 1 }{ 1+{ u }^{ 2 } } \cdot 2{ x }\\ &=\frac { 2x }{ 1+{ x }^{ 4 } }. \ _\square \end{aligned}$$

Find the derivative of $\displaystyle\int_{1}^{x^2}\cos t \, dt.$

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Again, we use the chain rule along with the fundamental theorem of calculus to solve this.

Let $u=x^2$, then

$$\begin{aligned} \displaystyle\int_{1}^{x^2}\cos t\, dt&=\dfrac{d}{du}\left[\displaystyle\int_{1}^{u}\cos t\, dt\right]\cdot\dfrac{du}{dx}\\ &=\cos{u}\cdot\dfrac{d}{dx}\left(x^2\right)\\ &=\cos u\cdot 2x\\ &=2x\cos x^2. \ _\square \end{aligned}$$

Find the derivative of $\displaystyle h(x)=\int _{ x }^{ 3x }{ \sin\theta \, d\theta }$.

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We use the following property of integrals:

$$\int _{ b }^{ a }{ f(x)\, dx } =\int _{ b }^{ 0 }{ f(x)\, dx+\int _{ 0 }^{ a }{ f(x)\, dx } } .$$

So,

$$\begin{aligned} h'(x) &=\frac { d }{ dx } \int _{ x }^{ 3x }{ \sin\theta \, d\theta } \\ &=\frac { d }{ dx } \int _{ x }^{ 0 }{ \sin\theta \, d\theta +\frac { d }{ dx } \int _{ 0 }^{ 3x }{ \sin\theta \, d\theta } } \\ &=\frac { d }{ dx } -\int _{ 0 }^{ x }{ \sin\theta \, d\theta } +\frac { d }{ dx } \int _{ 0 }^{ 3x }{ \sin\theta \, d\theta } \\ &=-\sin x+\frac { d }{ du } \int _{ 0 }^{ u }{ \sin\theta \, d\theta } \cdot \frac { du }{ dx } \\ &=-\sin x+3\sin3x. \ _\square \end{aligned}$$

If $f$ is a continuous function on $[a,b]$, then

$$\int _{ a }^{ b }{ f(x)\, dx=F(b)-F(a) } ,$$

where $F$ is an anti-derivative of $f$, i.e. $F'=f$. $_\square$

We know from the first fundamental theorem of calculus that if $\displaystyle S(x)=\int _{ a }^{ x }{ f(t)\, dt }$, then $S$ is an anti-derivative of $f$ or $S'(x)=f(x)$. And since $F'(x)=f(x),$

$$S'(x)=F'(x).$$

Integrating both sides with respect to $x$, we have

$$\displaystyle S(x)=\int _{ a }^{ x }{ f(t)\, dt }=F(x)+C, \qquad \qquad(1)$$

where $C$ is some constant.

Now, plugging in $x=a$ in this equation, we have

$$S(a)=\int _{ a }^{ a }{ f(t)\, dt }=F(a)+C.$$

By definition, $\displaystyle S(a)=\int _{ a }^{ a }{ f(t)\, dt }=0$, and hence

$$\begin{aligned} F(a)+C&=0\\ \Rightarrow C&=-F(a). \end{aligned}$$

Plugging in $C=-F(a)$ back in the equation $(1)$, we have

$$\displaystyle S(x)=\int _{ a }^{ x }{ f(t)\, dt }=F(x)-F(a).$$

Finally, setting $x=b$, we have

$$\begin{aligned} S(b)=\int _{ a }^{ b }{ f(t)\, dt }&=F(b)-F(a)\\ \Rightarrow \int _{ a }^{ b }{ f(x)\, dx }&=F(b)-F(a).\ _\square \end{aligned}$$

We could also prove the above theorem using the concept of Riemann sums.

We know from the first fundamental theorem of calculus that if $\displaystyle S(x)=\int _{ a }^{ x }{ f(t)\, dt }$, then $S$ is an anti-derivative of $f$ or $S'(x)=f(x)$. So we know that $\displaystyle f(x) = F'(x)$. Consider a partition $P = \{ a= x_0 < x_1 < x_2 < \dots < x_n = b \}$. With this partition as reference, we can write

$$F(b) - F(a) = \sum_{i=0}^{n-1} \big[F(x_{i+1}) - F(x_i)\big].$$

This is now the neater part of the proof. $\displaystyle \forall i = 0,1,2, \dots, n -1 \text{ } \exists t_i \in \text{ } ( x_i , x_{i+1} )$ such that $\displaystyle F(x_{i+1}) - F(x_i) = F'(t_i) \left( x_{i+1} - x_{i} \right) = f(t_i) \left( x_{i+1} - x_{i} \right)$. This is direct implication of the mean value theorem. So now what we have is

$$F(b) - F(a) = \sum_{i=0}^{n-1} f(t_i) \left( x_{i+1} - x_{i} \right);~ t_i \in \text{ } ( x_i , x_{i+1} ).$$

The right-hand side of the expression is nothing but the Riemann sum which will eventually converge to definite integral as the partition $P$ gets finer and finer:

$$F(b)-F(a)=\int _{ a }^{ b }{ f(t)\, dt } . \ _\square$$

This theorem transforms the difficult problem of evaluating definite integrals by calculating limits of sums, into an easier problem of finding an anti-derivative. So for example if we are asked to compute the integral $\displaystyle \int _{ a }^{ b }{ f(x)\, dx }$, we find an anti-derivative of $f(x)$ and compute their value at each end-point of the integral, and finally subtract them from each other.

Evaluate $\displaystyle{\int_0^1}x^2\, dx.$

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According to the fundamental theorem of calculus, we have

$$\displaystyle{\int_0^1}x^2\, dx=F(1)-F(0),$$

where $F(x)$ is an anti-derivative of $x^2.$ Indefinite integration of $x^2$ gives

$$\int x^2dx=\frac{1}{3}x^3+C,$$

where $C$ is the constant of integration.

Hence we have

$$F(1)-F(0)=\left(\frac{1}{3}\times1^3+C\right)-\left(\frac{1}{3}\times0+C\right)=\frac{1}{3}.\ _\square$$

Observe that the constant of integration $C$ can be ignored since it is to be eliminated regardless of its value.

Find the area under the curve $y=x^3+1$ from $x=-1$ to $x=1.$

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Since $x^3+1\geq0$ over the interval $[-1,1],$ the area under the curve $y=x^3+1$ from $x=-1$ to $x=1$ is equal to $\displaystyle{\int_{-1}^1\left(x^3+1\right)\, dx}.$ Since $\frac{1}{4}x^4+x$ is an anti-derivative of $x^3+1,$ the area under the curve is

$$\begin{aligned} \int_{-1}^1\left(x^3+1\right)\, dx&=\left[\frac{1}{4}x^4+x\right]_{-1}^1\\ &=\left(\frac{1}{4}\cdot1^4+1\right)-\left(\frac{1}{4}\cdot(-1)^4+(-1)\right)\\ &=\frac{5}{4}-\left(-\frac{3}{4}\right)\\ &=2.\ _\square \end{aligned}$$

Evaluate $\displaystyle{\int_{-3}^{2}\left(2x^2-3x+4\right)\, dx.}$

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Since $\displaystyle{\int\left(2x^2-3x+4\right)\, dx=\frac{2}{3}x^3-\frac{3}{2}x^2+4x+C,}$ we have

$$\begin{aligned} \displaystyle{\int_{-3}^{2}\left(2x^2-3x+4\right)\, dx}&=\left[\frac{2}{3}x^3-\frac{3}{2}x^2+4x\right]_{-3}^{2}\\ &=\left(\frac{2}{3}\cdot2^3-\frac{3}{2}\cdot2^2+4\cdot2\right)-\left(\frac{2}{3}\cdot(-3)^3-\frac{3}{2}\cdot(-3)^2+4\cdot(-3)\right)\\ &=\frac{305}{6}.\ _\square \end{aligned}$$

Evaluate $\displaystyle{\int_{4}^{9}\sqrt{x}\, dx.}$

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We have

$$\begin{aligned} \int_{4}^{9}\sqrt{x}\, dx&=\int_4^9 x^{\frac{1}{2}}\, dx\\ &=\left[\frac{2}{3}x^{\frac{3}{2}}\right]_4^9\\ &=\frac{2}{3}\cdot9^{\frac{3}{2}}-\frac{2}{3}\cdot4^{\frac{3}{2}}\\ &=\frac{38}{3}.\ _\square \end{aligned}$$

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What is the area of the shaded region in the above figure?

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We need to find the area under the curve $y=\frac{1}{x}$ from $x=\frac{1}{2}$ to $x=\frac{5}{2}.$ Since $\displaystyle{\int\frac{1}{x}\, dx=\ln x+C,}$ the area under the curve is equal to

$$\begin{aligned} \int_{\frac{1}{2}}^{\frac{5}{2}}\frac{1}{x}\, dx&=\Big[\ln x\Big]_{\frac{1}{2}}^{\frac{5}{2}}\\ &=\ln\frac{5}{2}-\ln\frac{1}{2}\\ &=\ln5.\ _\square \end{aligned}$$

Evaluate $\displaystyle\int_{-\frac{\pi}{4}}^{0} \sec x\tan x \ dx.$

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We know that

$$\dfrac{d}{dx}\sec x = \sec x\tan x.$$

Thus, the integral would be

$$\begin{aligned} \displaystyle\int_{-\frac{\pi}{4}}^{0} \sec x\tan x \ dx &= \Big[\sec x\Big]_{-\frac{\pi}{4}}^{0} \\ &=\sec 0 - \sec\left(\dfrac{-\pi}4\right)\\ &=1-\sqrt 2. \ _\square \end{aligned}$$

Evaluate $\displaystyle{\int_{-\frac{\pi}{2}}^{\frac{5}{6}\pi}\sin x\, dx.}$

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We have

$$\begin{aligned} \int_{-\frac{\pi}{2}}^{\frac{5}{6}\pi}\sin x\, dx&=\Big[-\cos x\Big]_{-\frac{\pi}{2}}^{\frac{5}{6}\pi}\\ &=\frac{\sqrt{3}}{2}.\ _\square \end{aligned}$$

Find the area under the curve $y=2x-x^2$ from $x=1$ to $x=2.$

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Since $2x-x^2\geq0$ over the interval $[1,2],$ the area under the curve $y=2x-x^2$ from $x=1$ to $x=2$ is equal to $\displaystyle{\int_1^2\left(2x-x^2\right)\, dx}.$ Since $x^2-\frac{1}{3}x^3$ is an anti-derivative of $2x-x^2,$

$$\begin{aligned} \int_1^2\left(2x-x^2\right)\, dx&=\left[x^2-\frac{1}{3}x^3\right]_1^2\\ &=\left(2^2-\frac{1}{3}\cdot2^3\right)-\left(1^2-\frac{1}{3}\cdot1^3\right)\\ &=\frac{2}{3}.\ _\square \end{aligned}$$