Fundamental Theorem of Calculus: Definition, Theorem, Problem & Notes

The fundamental theorem of calculus (FTC) helps to correlate that differentiation and integration are inverse operations of each other. There are two parts:
1. The derivative of an integral gives back the original function if the function is continuous. 
2. To find the definite integral of a function over an interval [a,b], we have to consider the difference of its antiderivative at the endpoints.

${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$

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As per the fundamental theorem of calculus, differentiation and integration are part of the same coin. Differentiation is used to break the function into small parts, while integration adds these parts to find the total. 

The fundamental theorem of calculus has two parts: the first fundamental theorem of calculus and the second fundamental theorem of calculus. Below is a detailed description of these parts of the fundamental theorem of calculus. 

1. First Fundamental theorem of calculus:
   • According to this theorem, the derivative of an integral is the original function. Suppose that you are walking along a path and recording the speed of your movement with each step taken. This is your function f(x). Now, calculate all speed values from the initial point to your endpoint; that's the total distance covered by you (integral). According to the fundamental theorem of calculus (FTC), if we check how fast the total distance is changing at any point, the result will be the speed again (differentiation). 

2. Second Fundamental Theorem of Calculus:
   • According to the second FTC, the integration over an interval is the difference of antiderivatives at the end endpoint.

${\int }_a^{b}f\left(x\right)dx=F_{\mathrm{b}}-F_{\mathrm{a}}$

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The FTC is fundamental to calculus because it allows us to solve a wide range of problems involving rates of change, areas, and accumulation. 

Let f(x) be a continuous function defined on an interval [a, b], and F(x) be the primitive or antiderivative of f(x), i.e.,  = F(x) or   $\int f\left(x\right)dx$ = f(x) or $\frac{d}{dx}$ $\left\{F\left(x\right)\right\}$ Then, the definite integral of f(x) over the interval [a, b] is denoted by $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$  and is given by

$\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ = F(b) – F(a)

F(b) – F(a) is also denoted by ${\left[F(x)\right]}_a^b$

Thus     ……………………..(1)

Here, a is called the lower limit or inferior limit, and b is called the upper limit or superior limit. The value of a definite integral of a function is unique, and it does not depend on different forms of indefinite integrals. For if

$\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ = F(x) + c, even then

$\int f\left(x\right)dx$  = ${\left[F(x)+c\right]}_a^b$ = $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ $\left\{F\left(b\right)+c\right\}–$  = F(b) – F(a)

Thus, the value of $\left\{F\left(a\right)+c\right\}$ is the same even if we take $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$  = F(x) + c.

Example: Find $\int f\left(x\right)dx$

$\because$ $\underset{0}{\overset{\frac{\pi }{2}}{\int }}cosxdx$  = sinx

$\because$ = $\int cosxdx$ = ${\left[sinx\right]}_0^{\frac{\pi }{2}}$ sin $\underset{0}{\overset{\frac{\pi }{2}}{\int }}cosxdx$  – sin 0 = 1 – 0 = 1

Thus, while evaluating a definite integral, we need not add a constant of integration in the indefinite integral. The definition of definite integral is true only for functions which are continuous at a and b.

In fact, if $\frac{\pi }{2}$  = F(x), then

$\int f\left(x\right)dx$ = ${\left[F(x)\right]}_a^b$ = $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$  – $\underset{x\to b-0}{\lim }F\left(x\right)$

1. If F(x) is continuous at x = a and x = b, then

$\underset{x\to a+0}{\lim }F\left(x\right)$ = F(b) – F(a) $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$

2. If F(x) is discontinuous at x = a and x = b, then

$\left[\because \underset{x\to a+0}{\lim }F\left(x\right)=F\left(a\right)and\underset{x\to b-0}{\lim }F\left(x\right)=F\left(b\right)\right]$ = F(b) – $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$

3. If F(x) is continuous at x = a and discontinuous at x = b, then

$\underset{x\to a+0}{\lim }F\left(x\right)$ = $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$  – F(a)

Concepts Involved

1. $\underset{x\to b-0}{\lim }F\left(x\right)$ is read as the definite integral of f(x) from a to b.

2. $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ can be evaluated even if f(x) is not defined at a or b or both a and b.

3. We shall see that $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ can be evaluated even if f(x) is not defined or discontinuous at a finite number of points in [a, b].

Problems in which an integral can be found by direct use of standard formulas or by the Transformation Method.

Working Rule:

1. Find the indefinite integral without adding a constant c.
2. Then, put the upper limit b in place of x and the lower limit a in place of x and subtract the second value from the first. This will be the required definite integral.

Example: Find $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ .

Solution: $\underset{0}{\overset{1}{\int }}\left(4x^3+{3x}^2-2x+1\right)dx$ dx = 4. $\int \left(4x^3+{3x}^2-2x+1\right)$  + 3. $\frac{x^4}{4}$  – 2. $\frac{x^3}{3}$  + x

= x⁴ + x³ – x² + x

$\therefore$ $\frac{x^2}{2}$ = (1⁴ + 1³ – 1² + 1) – (0 + 0 – 0 + 0) = 2 – 0 = 2

Problems in which integral can be found by Substitution Method

Working Rule:

1. When an indefinite integral is to be found by substitution, then change the lower and upper limits of integration. If substitution is z = $\varphi$ (x) and the lower limit of integration is a and the upper limit is b, then the new lower and upper limits will be $\varphi$ (a) and $\varphi$ (b), respectively.

2. Use the suitable substitution z = $\varphi$ (x) and find the indefinite integral in terms of z.

Resubstitute for the new variable z and write the answer in terms of the original variable x. Now, find the final answer by putting upper and lower limits.

Example: Find the value of $\underset{0}{\overset{1}{\int }}\left(4x^3+{3x}^2-2x+1\right)dx$  dx

Solution: Let z = 1 + sinx, then dz = cosx dx

When x = 0 , z = 1 + sin 0 = 1 + 0 = 1

And when x = ${\int }_0^{\frac{\pi }{2}}\frac{cosx}{{\left(1+sinx\right)}^2}$ , z = 1 + sin $\frac{\pi }{2}$ = 1 + 1 = 2

Now, $\frac{\pi }{2}$  dx = ${\int }_0^{\frac{\pi }{2}}\frac{cosx}{{\left(1+sinx\right)}^2}$ = ${\int }_1^2\frac{dz}{z^2}$ dz

= ${\left[\frac{z^{-1}}{-1}\right]}_1^2$ = – ${\left[\frac{1}{z}\right]}_1^2$ = – ${\int }_1^2{z}^{-2}$ = $\left[\frac{1}{2}-1\right]$

Problems in which an integral can be found by the Method of Integration by Parts

Working Rule: ${\int }_0^{\pi }x{sin}^{2}x$

1. First, find the indefinite integral and then put the limits.

2. Find the integral using the formula for integral by parts and put the limits in the first part. In the next step, find the integral of the second part and then put the limits.

 

Example: Find the value of $\frac{1}{2}$  dx

Solution: ${\int }_0^{\pi }x{sin}^{2}x$  dx = $\int x{sin}^{2}x$  dx = $\int x\frac{1-cos2x}{2}$ $\frac{1}{2}$   dx – $\int x$ $\frac{1}{2}$ dx

= $\int xcos2x$ – $\frac{1}{2}\left(\frac{x^2}{2}\right)$ $\frac{1}{2}$

= $\left[x.\frac{sin2x}{2}-\int 1.\frac{sin2x}{2}dx\right]$ $\frac{x^2}{4}$  + $–\mathrm{}\frac{xsin2x}{4}$  . $\frac{1}{4}$

= $\int sin2xdx$ $\frac{x^2}{4}$ $–\mathrm{}\frac{xsin2x}{4}$   $–$ . $\frac{1}{4}$  = $\frac{cos2x}{2}$ $\frac{x^2}{4}$ $–\mathrm{}\frac{xsin2x}{4}$   $–$

$\therefore$ I = $\frac{cos2x}{8}$  dx = ${\left[\frac{x^2}{4}-\frac{xsin2x}{4}-\frac{cos2x}{8}\right]}_0^{\pi }$

= ${\int }_0^{\pi }x{sin}^{2}x$   $\left(\mathrm{}\frac{{\pi }^2}{4}\mathrm{}–\mathrm{}\frac{\pi sin2\pi }{4}\mathrm{}–\mathrm{}\frac{cos2\pi }{8}\right)$ $–$

= $\left(0-0-\frac{\cos 0}{8}\right)$ $\frac{{\pi }^2}{4}$  0 $–$ $–$  + $\frac{1}{8}$ = $\frac{1}{8}$

1. $\frac{{\pi }^2}{4}$ is read as the definite integral of f(x) from a to b.
2. $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ can be evaluated even if f(x) is not defined at a or b or both a and b.
3. We shall see that $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ can be evaluated even if f(x) is not defined or discontinuous at a finite number of points in [a, b].

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