Using Riemann sums to calculate definite integrals can be tedious, as was seen in the previous section. In fact the technique shown in that section depended on the function being a low-degree polynomial, which obviously will not always be the case. Luckily there is a better way, involving antiderivatives, given by the following theorem:
The function \(A(x)\) in Part I of the theorem is sometimes called the area function because it represents the area under the curve \(y=f(x)\) over the interval \(\ival{a}{x}\), as shown in Figure [fig:ftc1] below.
To prove Part I, assume that \(f(x) \ge 0\) on \(\ival{a}{b}\) as in Figure [fig:ftc1] (the proofs for \(f(x)\) either negative or switching sign over \(\ival{a}{b}\) are similar). The goal is to show that for any \(x\) in \(\ival{a}{b}\) the differential \(\dA\) exists and equals \(f(x)\,\dx\). First, \(\dA = A(x+\dx) - A(x)\) is the area under the curve \(y=f(x)\) over the interval \(\ival{x}{x+\dx}\), as shown in Figure [fig:ftc1dA] above.
By the Microstraightness Property \(f\) is a straight line over the infinitesimal interval \(\ival{x}{x+\dx}\), so \(f\) must be either increasing, constant, or decreasing over that interval. The three possibilities are shown in Figure [fig:ftc1dA3]:
In the case where \(f\) is increasing over \(\ival{x}{x+\dx}\), the infinitesimal area \(\dA\) is the sum of the area of the rectangle of height \(f(x)\) and width \(\dx\) and the area of the right triangle \(\triangle ABC\) shown in Figure [fig:ftc1dA3](a). The area of \(\triangle ABC\) is \(\frac{1}{2}(\df)(\dx) = \frac{1}{2}f'(x)(\dx)^2 = 0\), so \(\dA = f(x)\,\dx\).
In the case where \(f\) is constant over \(\ival{x}{x+\dx}\), the infinitesimal area \(\dA\) is the area of the rectangle of height \(f(x)\) and width \(\dx\), as shown in Figure [fig:ftc1dA3](b). So again, \(\dA = f(x)\,\dx\).
In the case where \(f\) is decreasing over \(\ival{x}{x+\dx}\), the infinitesimal area \(\dA\) is the sum of the area of the rectangle of height \(f(x+\dx)\) and width \(\dx\) and the area of the right triangle \(\triangle ABC\) shown in Figure [fig:ftc1dA3](c). Note that \(\df < 0\) since \(f\) is decreasing, and so the area of \(\triangle ABC\) is \(\frac{1}{2}(-\df)(\dx) = -\frac{1}{2}f'(x)(\dx)^2 = 0\). Thus,
\[\dA ~=~ f(x+\dx)\,\dx ~=~ (f(x) + \df)\,\dx ~=~ f(x)\,\dx ~+~ f'(x)(\dx)^2 ~=~ f(x)\,\dx ~+~ 0 ~=~ f(x)\,\dx ~. \nonumber \]
So in all three cases, \(\dA = f(x)\,\dx\), and so \(A'(x) = \frac{\dA}{\dx} = f(x)\), which shows that \(A(x)\) is differentiable and has derivative \(f(x)\). This proves Part I of the Fundamental Theorem of Calculus.\(\quad\checkmark\)
To prove Part II of the theorem, let \(F(x)\) be an antiderivative of \(f(x)\) over \(\ival{a}{b}\). Since \(A(x) = \int_a^x f(x)\,\dx\) is also an antiderivative of \(f(x)\) over \(\ival{a}{b}\) by Part I of the theorem, then \(A(x)\) and \(F(x)\) differ by a constant \(C\) over \(\ival{a}{b}\). In other words:
\[A(x) ~=~ F(x) ~+~ C \quad\text{for all $x$ in $\ival{a}{b}$} \nonumber \]
By definition \(A(a) = 0\), since it is the area under the curve over the interval \(\ival{a}{a}\) of zero length. Thus,
\[0 ~=~ A(a) ~=~ F(a) ~+~ C \quad\Rightarrow\quad C ~=~ -F(a) \quad\Rightarrow\quad A(x) ~=~ F(x) ~-~ F(a) \quad\text{for all $x$ in $\ival{a}{b}$} \nonumber \]
and so
\[\int_a^b f(x)~\dx ~=~ A(b) ~=~ F(b) ~-~ F(a) \nonumber \]
which proves Part II of the theorem.3 \(\quad\checkmark\)
Note: In some textbooks Part I is called the First Fundamental Theorem of Calculus and Part II is called the Second Fundamental Theorem of Calculus. The following notation provides a shorthand way of writing \(F(b) - F(a)\):
Example \(\PageIndex{1}\): ftc1
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Solution
Calculate \(\displaystyle\int_1^2 x^2~\dx\).
Solution: Recall from Example
Example \(\PageIndex{1}\): riemann1
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Solution
in the previous section that the integral equals \(7/3\). In that example Riemann sums were used, but Part II of the Fundamental Theorem of Calculus makes the integral much easier to calculate. Since \(F(x) = \frac{x^3}{3}\) is an antiderivative of \(f(x) = x^2\), then
\[\int_1^2 x^2~\dx ~=~ \frac{x^3}{3}~\Biggr|_1^{2} ~=~ \frac{2^3}{3} ~-~ \frac{1^3}{3} ~=~ \frac{7}{3} ~. \nonumber \]
Note in the above example that any antiderivative of \(f(x)=x^2\) could have been used, e.g. \(F(x)= \frac{x^3}{3} + 5\). Notice that the constant 5 would have been canceled out when evaluating \(F(2)-F(1)\). So you do not need to add a generic constant \(C\) to the antiderivative of \(f(x)\) in a definite integral, as you would in an indefinite integral.
Example \(\PageIndex{1}\): ftc2
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Solution
Calculate \(\displaystyle\int_0^{\pi} \sin\,x~\dx\).
Solution: Since \(F(x) = -\cos\,x\) is an antiderivative of \(f(x) = \sin\,x\), then
\[\int_0^{\pi} \sin\,x~\dx ~=~ -\cos\,x~\Biggr|_0^{\pi} ~=~ -\cos\,\pi ~-~ (-\cos\,0) ~=~ -(-1) ~-~ (-1) ~=~ 2 ~. \nonumber \]
Example \(\PageIndex{1}\): ftc3
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Solution
Calculate \(\displaystyle\int_{-1}^{1} x^3~\dx\).
Solution: Since \(F(x) = \frac{x^4}{4}\) is an antiderivative of \(f(x) = x^3\), then
\[\int_{-1}^{1} x^3~\dx ~=~ \frac{x^4}{4}~\Biggr|_{-1}^{1} ~=~ \frac{1^4}{4} ~-~ \frac{(-1)^4}{4} ~=~ \frac{1}{4} ~-~ \frac{1}{4} ~=~ 0~. \nonumber \]
Example
Example \(\PageIndex{1}\): ftc3
is a special case of the following result for odd functions:
The idea is that since an odd function is symmetric around the origin, then the area between the curve and the \(x\)-axis over \(\ival{0}{a}\) will cancel out the area between the curve and the \(x\)-axis over \(\ival{-a}{0}\). Both areas are the same but one gets counted as positive and the other negative, as shown in Figure [fig:defintodd] below:
By symmetry around the \(y\)-axis, a similar result holds for even functions (see Figure [fig:definteven]):
The following rules for definite integrals are a consequence of the corresponding rules for indefinite integrals:
The following results for definite integrals are a consequence of the Fundamental Theorem of Calculus:
For example, if \(F(x)\) is an antiderivative of \(f(x)\) on \(\ival{a}{b}\), then
\[\int_a^c f(x)~\dx ~+~ \int_c^b f(x)~\dx ~=~ (F(c) - F(a)) ~+~ (F(b) - F(c)) ~=~ F(b) ~-~ F(a) ~=~ \int_a^b f(x)~\dx \nonumber \]
which proves rule (3).
The following result is a consequence of Part I of the Fundamental Theorem of Calculus along with the Chain Rule:
Example \(\PageIndex{1}\): ftc4
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Solution
Let \(F(x) = \displaystyle\int_0^{x^2} e^{-t^2}~\dt\) for all \(x > 0\). Find \(F'(x)\).
Solution: By the Chain Rule for integrals, with \(f(t) = e^{t^2}\) and \(g(x) = x^2\):
\[F'(x) ~=~ f(g(x))\cdot g'(x) ~=~ e^{-(x^2)^2} \cdot (2x) ~=~ 2x\,e^{-x^4} \nonumber \]
[sec5dot3]
For Exercises 1-12, evaluate the given definite integral.
4
\(\displaystyle\int_{0}^{1}\, x^2 ~\dx\)
\(\displaystyle\int_{-1}^{1}\, x^2 ~\dx\)
\(\displaystyle\int_{0}^{1}\, x^3 ~\dx\)
\(\displaystyle\int_{-1}^{1}\, (x^2 + 3x - 4) ~\dx\)
4
\(\displaystyle\int_{1}^{2}\, \frac{1}{x^2} ~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}\)
\(\displaystyle\int_{2}^{3}\, \frac{1}{x^3} ~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}\)
\(\displaystyle\int_{0}^{\pi/2}\, \cos\,x ~\dx\)
\(\displaystyle\int_{0}^{1}\, e^x ~\dx\vphantom{\displaystyle\int_{0}^{\pi/2}}\)
4
\(\displaystyle\int_{-1}^{1}\, 2e^x ~\dx\vphantom{\dfrac{x^3\,e^{x^2}}{\cos\,2x}}\)
\(\displaystyle\int_{-\pi}^{\pi}\, \sin\,x ~\dx\vphantom{\dfrac{x^3\,e^{x^2}}{\cos\,2x}}\)
\(\displaystyle\int_{0}^{4}\, \sqrt{x} ~\dx\vphantom{\dfrac{x^3\,e^{x^2}}{\cos\,2x}}\)
\(\displaystyle\int_{-2}^{2}\, \dfrac{x^3\,e^{x^2}}{\cos\,2x} ~\dx\)
2
Show that \(\ln\,x ~=~ \displaystyle\int_{1}^{x}\, \frac{1}{t} ~\dt\) for all \(x > 0\).
Show that \(\displaystyle\int_b^a\,f(x)~\dx ~=~ -\displaystyle\int_a^b\,f(x)~\dx\).
Given \(f(x) ~=~\displaystyle\int_{1}^{x}~\dfrac{t}{\sqrt{t^4 + 1}}~\dt~\), find \(f'(3)\) and \(f'(-2)\). [[1.]]
Prove the Chain Rule for integrals.
Explain why for any continuous function \(f\) on \(\ival{a}{b}\), \(\ABS{\displaystyle\int_a^b\,f(x)~\dx} ~\le~ \displaystyle\int_a^b\,\abs{f(x)}~\dx ~.\)
Explain why if \(f(x) \le g(x)\) on \(\ival{a}{b}\) then \(\displaystyle\int_a^b\,f(x)~\dx ~\le~ \displaystyle\int_a^b\,g(x)~\dx ~.\) [[1.]]
Show that if \(f(x)\) is continuous on \(\ival{a}{b}\) then there is a number \(c\) in \((a,b)\) such that
\[\int_a^b f(x)~\dx = f(c) \cdot (b-a) ~. \nonumber \]
Let \(f(t)\) be a continuous function for all \(t \ge 0\), and for each \(x \ge 0\) define a function \(g(x)\) by
\[g(x) ~=~ \displaystyle\int_0^x ~(x-t)\,f(t)~\dt ~. \nonumber \]
Show that \(g'(x) ~=~ \displaystyle\int_0^x ~f(t)~\dt ~\) for all \(x \ge 0\).
Show that for all \(x > 0\),
\[\int_0^x \frac{\dt}{1+t^2} ~+~ \int_0^{1/x} \frac{\dt}{1+t^2} \nonumber \]
is independent of \(x\).
