The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) shows that differentiation and integration are inverse processes.

Part 1 (FTC1)

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

\[g\left( x \right) = \int\limits_a^x {f\left( {t} \right)dt},\;\; a \le x \le b\]

is an antiderivative of f, that is

\[g^\prime\left( x \right) = f\left( x \right)\;\;\text{or}\;\;\frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) = f\left( x \right).\]

If f happens to be a positive function, then g (x) can be interpreted as the area under the graph of f from a to x.

The Fundamental Theorem of Calculus (Part 1) — Figure 1.

The first part of the theorem says that if we first integrate \(f\) and then differentiate the result, we get back to the original function \(f.\)

Part 2 (FTC2)

The second part of the fundamental theorem tells us how we can calculate a definite integral.

If \(f\) is a continuous function on \(\left[ {a,b} \right]\) and \(F\) is an antiderivative of \(f,\) that is \(F^\prime = f,\) then

\[\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)\;\;\text{or}\;\;\int\limits_a^b {{F^\prime\left( x \right)}dx} = F\left( b \right) - F\left( a \right).\]

To evaluate the definite integral of a function \(f\) from \(a\) to \(b,\) we just need to find its antiderivative \(F\) and compute the difference between the values of the antiderivative at \(b\) and \(a.\)

So the second part of the fundamental theorem says that if we take a function \(F,\) first differentiate it, and then integrate the result, we arrive back at the original function, but in the form \(F\left( b \right) - F\left( a \right).\)

Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes.

The Area under a Curve and between Two Curves

The area under the graph of the function \(f\left( x \right)\) between the vertical lines \(x = a,\) \(x = b\) (Figure \(2\)) is given by the formula

\[S = \int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right).\]

Let \(F\left( x \right)\) and \(G\left( x \right)\) be antiderivatives of functions \(f\left( x \right)\) and \(g\left( x \right),\) respectively.

If \(f\left( x \right) \ge g\left( x \right)\) on the closed interval \(\left[ {a,b} \right],\) then the area between the curves \(y = f\left( x \right),\) \(y = g\left( x \right)\) and the lines \(x = a,\) \(x = b\) (Figure \(3\)) is given by

\[S = \int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = F\left( b \right) - G\left( b \right) - F\left( a \right) + G\left( a \right).\]

The Method of Substitution for Definite Integrals

The definite integral \(\int\limits_a^b {f\left( x \right)dx} \) of the variable \(x\) can be changed into an integral with respect to \(t\) by making the substitution \(x = g\left( t \right):\)

\[\int\limits_a^b {f\left( x \right)dx} = \int\limits_c^d {f\left( {g\left( t \right)} \right)g'\left( t \right)dt} .\]

The new limits of integration for the variable \(t\) are given by the formulas

\[c = {g^{ - 1}}\left( a \right),\;\;d = {g^{ - 1}}\left( b \right),\]

where \({g^{ - 1}}\) is the inverse function to \(g,\) that is \(t = {g^{ - 1}}\left( x \right).\)

Integration by Parts for Definite Integrals

In this case the formula for integration by parts looks as follows:

\[\int\limits_a^b {udv} = \left. {uv} \right|_a^b - \int\limits_a^b {vdu} ,\]

where \(\left. {uv} \right|_a^b\) means the difference between the product of functions \(uv\) at \(x = b\) and \(x = a.\)

Solved Problems

Example 1.

Calculate the derivative of the function \[g\left( x \right) = \int\limits_1^x {\sqrt {{t^3} + 4t} dt} \] at \(x = 2.\)

Solution.

We apply the Fundamental Theorem of Calculus, Part \(1:\)

\[g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) = f\left( x \right).\]

Hence

\[g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_1^x {\sqrt {{t^3} + 4t} dt} } \right) = \sqrt {{x^3} + 4x} .\]

Substituting \(x = 2\) yields

\[g^\prime\left( 2 \right) = \sqrt {{2^3} + 4 \cdot 2} = \sqrt {16} = 4.\]

Example 2.

Calculate the derivative of the function \[g\left( x \right) = \int\limits_{ - \frac{\pi }{2}}^x {\sqrt {{{\sin }^2}t + 2} dt} \] at \(x = \frac{\pi }{6}.\)

Solution.

We use the Fundamental Theorem of Calculus, Part \(1:\)

\[g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_a^x {f\left( t \right)dt} } \right) = f\left( x \right).\]

Then

\[g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_{ - \frac{\pi }{2}}^x {\sqrt {{{\sin }^2}t + 2} dt} } \right) = \sqrt {{{\sin }^2}x + 2} .\]

Note that the lower limit of integration \({ - \frac{\pi }{2}}\) does not affect the answer.

Now we compute the value of the derivative for \(x = \frac{\pi }{6} :\)

\[g^\prime\left( {\frac{\pi }{6}} \right) = \sqrt {{{\sin }^2}\frac{\pi }{6} + 2} = \sqrt {{{\left( {\frac{1}{2}} \right)}^2} + 2} = \sqrt {\frac{9}{4}} = \frac{3}{2}.\]

Example 3.

Find the derivative of the function \[g\left( x \right) = \int\limits_3^{{x^2}} {\frac{{dt}}{t}}.\]

Solution.

We introduce the new function

\[h\left( u \right) = \int\limits_3^u {\frac{{dt}}{t}}.\]

Using the FTC1, we have

\[h^\prime\left( u \right) = \frac{1}{u}.\]

As \(g\left( x \right) = h\left( {{x^2}} \right),\) then by the chain rule

\[g^\prime\left( x \right) = \left[ {h\left( {{x^2}} \right)} \right]^\prime = h^\prime\left( {{x^2}} \right) \cdot \left( {{x^2}} \right)^\prime = h^\prime\left( {{x^2}} \right) \cdot 2x = \frac{1}{{{x^2}}} \cdot 2x = \frac{2}{x}.\]

Example 4.

Find the derivative of the function \[g\left( x \right) = \int\limits_0^{{x^2}} {\sqrt {1 + {t^2}} dt}.\]

Solution.

Since the upper limit of integration is not \(x,\) we apply the chain rule. Let \(u = {x^2},\) then \(u^\prime = 2x.\)

Consider the new function

\[h\left( u \right) = \int\limits_0^u {\sqrt {1 + {t^2}} dt} .\]

By the FTC1, we can write

\[h^\prime\left( u \right) = \sqrt {1 + {u^2}} .\]

As \(g\left( x \right) = h\left( {{x^2}} \right),\) we have

\[g^\prime\left( x \right) = \left[ {h\left( {{x^2}} \right)} \right]^\prime = h^\prime\left( {{x^2}} \right) \cdot \left( {{x^2}} \right)^\prime = \sqrt {1 + {{\left( {{x^2}} \right)}^2}} \cdot 2x = 2x\sqrt {1 + {x^4}} .\]

Example 5.

Find the derivative of the function \[g\left( x \right) = \int\limits_1^{{x^3}} {{t^2}dt}.\]

Solution.

Let \(u = {x^3},\) then \(u^\prime = 3{x^2}.\)

We introduce the new function

\[h\left( u \right) = \int\limits_0^u {{t^2}dt} .\]

Using the FTC1, we obtain

\[h^\prime\left( u \right) = {u^2}.\]

Since \(g\left( x \right) = h\left( {{x^3}} \right),\) we have

\[g^\prime\left( x \right) = \left[ {h\left( {{x^3}} \right)} \right]^\prime = h^\prime\left( {{x^3}} \right) \cdot \left( {{x^3}} \right)^\prime = {\left( {{x^3}} \right)^2} \cdot 3{x^2} = {x^6} \cdot 3{x^2} = 3{x^8}.\]

Example 6.

Find the derivative of the function \[g\left( x \right) = \int\limits_{{x^2}}^{{x^3}} {tdt}.\]

Solution.

We split the interval of integration \(\left[ {{x^2},{x^3}} \right]\) using an intermediate point \(c,\) so that \(c \in \left[ {{x^2},{x^3}} \right].\) Hence the derivative of \(g\left( x \right)\) is written in the form

\[g^\prime\left( x \right) = \frac{d}{{dx}}\left( {\int\limits_{{x^2}}^{{x^3}} {tdt} } \right) = \frac{d}{{dx}}\left( {\int\limits_{{x^2}}^c {tdt} + \int\limits_c^{{x^3}} {tdt} } \right) = \frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} - \int\limits_c^{{x^2}} {tdt} } \right) = \frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} } \right) - \frac{d}{{dx}}\left( {\int\limits_c^{{x^2}} {tdt} } \right).\]

We calculate both terms using the FTC1 and the chain rule:

\[\frac{d}{{dx}}\left( {\int\limits_c^{{x^3}} {tdt} } \right) = {x^3} \cdot \left( {{x^3}} \right)^\prime = {x^3} \cdot 3{x^2} = 3{x^5};\]

\[\frac{d}{{dx}}\left( {\int\limits_c^{{x^2}} {tdt} } \right) = {x^2} \cdot \left( {{x^2}} \right)^\prime = {x^2} \cdot 2x = 2{x^3}.\]

Then

\[g^\prime\left( x \right) = 3{x^5} - 2{x^3}.\]

Example 7.

Calculate the derivative of the function \[g\left( x \right) = \int\limits_{\sqrt x }^x {\left( {{t^2} - t} \right)dt} \] at \(x = 1.\)

Solution.

We split the integral function into two terms:

\[g\left( x \right) = \int\limits_{\sqrt x }^x {\left( {{t^2} - t} \right)dt} = \int\limits_{\sqrt x }^c {\left( {{t^2} - t} \right)dt} + \int\limits_c^x {\left( {{t^2} - t} \right)dt} = \int\limits_c^x {\left( {{t^2} - t} \right)dt} - \int\limits_c^{\sqrt x } {\left( {{t^2} - t} \right)dt},\]

where \(c \in \left[ {{x^2},{x^3}} \right].\)

Find the derivative of \(g\left( x \right)\) using the FTC1 and the chain rule (for the second term):

\[\frac{d}{{dx}}\int\limits_c^x {\left( {{t^2} - t} \right)dt} = {x^2} - x;\]

\[\frac{d}{{dx}}\int\limits_c^{\sqrt x } {\left( {{t^2} - t} \right)dt} = \left( {{{\left( {\sqrt x } \right)}^2} - \sqrt x } \right) \cdot \left( {\sqrt x } \right)^\prime = \left( {x - \sqrt x } \right) \cdot \frac{1}{{2\sqrt x }} = \frac{{\sqrt x }}{2} - \frac{1}{2}.\]

Then

\[g^\prime\left( x \right) = \left( {{x^2} - x} \right) - \left( {\frac{{\sqrt x }}{2} - \frac{1}{2}} \right) = {x^2} - x - \frac{{\sqrt x }}{2} + \frac{1}{2}.\]

At the point \(x = 1,\) the derivative is equal to

\[g^\prime\left( 1 \right) = {1^2} - 1 - \frac{{\sqrt 1 }}{2} + \frac{1}{2} = 0.\]

Example 8.

Evaluate the integral \[\int\limits_0^2 {\left( {{x^3} - {x^2}} \right)dx}.\]

Solution.

Using the Fundamental Theorem of Calculus, Part \(2,\) we have

\[\int\limits_0^2 {\left( {{x^3} - {x^2}} \right)dx} = \left. {\left( {\frac{{{x^4}}}{4} - \frac{{{x^3}}}{3}} \right)} \right|_0^2 = \left( {\frac{{16}}{4} - \frac{8}{3}} \right) - 0 = \frac{4}{3}.\]

See more problems on Page 2.