4.6 The Fundamental Theorem of Calculus

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In this section we learn to compute the value of a definite integral using the fundamental theorem of calculus.

The Fundamental Theorem of Calculus

We begin by recalling the definition of the definite integral: $\int _a^b f(x) \ dx \equiv \lim _{n\to \infty } \sum _{i=1}^n f(x_i^*)\Delta x.$ where $\Delta x = (b-a)/n$ and $x_i^*$ is a sample point for the $i$ -th sub-interval, $[x_{i-1}, x_i]$ .

Computing the value of a definite integral from this definition can be cumbersome and require knowledge of special summation formulas. Fortunately, there is a theorem which states that computing definite integrals can done via anti-differentiation!

Fundamental Theorem of Calculus If $f(x)$ is continuous on the interval $[a, b]$ and if $F(x)$ is an anti-derivative of $f(x)$ , i.e., $F'(x) = f(x)$ , then $\int _a^b f(x) \ dx = F(b) - F(a).$

The proof of the Fundamental Theorem of Calculus can be obtained by applying the Mean Value Theorem to $F(x)$ on each of the sub-intervals $[x_{i-1}, x_i]$ and using the value of $c$ in each case as the sample point.

The process of calculating the numerical value of a definite integral is performed in two main steps: first, find the anti-derivative $F(x)$ and second, plug the endpoints of integration, $a$ and $b$ to compute $F(b) - F(a)$ . To symbolize the steps, we use a vertical evaluation bar:

$\int _a^b f(x) \ dx = F(x)\Big |_a^b = F(b) - F(a)$ In the first step, the function $F(x)$ is introduced and in the second step, the difference of the values $F(b)$ and $F(a)$ is computed.

In the following examples, we compute definite integrals using the FTC in order to solve the area problem. Recall that the definite integral of a non-negative function gives the area under the curve.

Find the exact area under the graph of $y = x^2$ from $x = 0$ to $x = 1$ .
Since the function $x^2$ is non-negative on the interval $[0, 1]$ , the exact area under the curve is the definite integral $\int _0^1 x^2 \ dx.$ We can compute the value of this definite integral using the Fundamental Theorem of Calculus as follows: $\int _0^1 x^2 \ dx = \frac {x^3}{3} \Bigg |_0^1 = \frac {1^3}{3} - \frac {0^3}{3} = \frac 13.$

Find the exact area under the curve $y = \cos (x)$ from $x=0$ to $x= \pi /2$ .
Since the function $\cos (x)$ is non-negative on the interval $[0, \frac {\pi }{2}]$ , the exact area can be expressed as the definite integral $\int _0^{\pi /2} \cos (x) \ dx.$ We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows: $\int _0^{\pi /2} \cos (x) \ dx = \sin (x) \Big |_0^{\pi /2} = \sin (\tfrac {\pi }{2}) - \sin (0)= 1- 0 = 1.$

Find the exact area under the graph of $y = \sin (x)$ from $x = 0$ to $x = \pi$ .
The area is $\answer [given]{2}$

$\text {Area} = \int _0^{\pi } \sin (x) \ dx$

An anti-derivative of $\sin (x)$ is $-\cos (x)$

The Fundamental Theorem says: $\int _a^b f(x) \ dx = F(b) - F(a)$

$\cos (0) = 1$ and $\cos (\pi ) = -1$

Find the exact area under the curve $y = e^x$ from $x = 0$ to $x = 1$ .
Since the function $e^x$ is non-negative on the interval $[0, 1]$ , the area can be expressed as the definite integral $\int _0^1 e^x \ dx.$ We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows: $\int _0^{1} e^x \ dx = e^x \Big |_0^{1} = e^{1} - e^0= e- 1.$

Find the exact area under the graph of $y = e^x$ from $x=1$ to $x = \ln (5)$ .
The area is $\answer [given]{5-e}$ .

$\text {Area} = \int _1^{\ln (5)} e^x \ dx$

An anti-derivative of $e^x$ is itself

The Fundamental Theorem says: $\int _a^b f(x) \ dx = F(b) - F(a)$

$e^{\ln (a)} = a$ if $a > 0$

Find the exact area under the graph of $y = x^3$ from $x = 2$ to $x = 4$ .
The area is $\answer [given]{60}$ .

$\text {Area}=\int _2^{4} x^3 \ dx$

An anti-derivative of $x^3$ is $x^4 /4$

The Fundamental Theorem says: $\int _a^b f(x) \ dx = F(b) - F(a)$

Find the exact area under the curve $y = 1/x$ from $x=1$ to $x= e$ .
Since the function $1/x$ is non-negative on the interval $[1, e]$ , the exact area can be expressed as the definite integral $\int _1^{e} \frac {1}{x} \ dx.$ We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows: $\int _1^e \frac {1}{x} \ dx = \ln |x| \Big |_1^e = \ln (e) - \ln (1)= 1- 0 = 1.$

Find the exact area under the graph of $y = 1/x$ from $x = 2$ to $x = 5$ .
The area is $\answer [given]{\ln (5/2)}$ .

$\text {Area} = \int _2^{5} \frac {1}{x} \ dx$

An anti-derivative of $1/x$ is $\ln (x)$

The Fundamental Theorem says: $\int _a^b f(x) \ dx = F(b) - F(a)$

Find the exact area under the curve $y = 2x+1$ from $x=0$ to $x= 2$ .
Since the function $2x+1$ is non-negative on the interval $[0, 2]$ , the exact area can be expressed as the definite integral $\int _0^2 (2x+1) \ dx.$ We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows: $\int _0^2 (2x+ 1) \ dx = (x^2 + x) \Big |_0^2 = (2^2 + 2) - (0^2 + 0) = 6.$

Find the exact area under the graph of $y = 4x-3$ from $x = 1$ to $x = 3$ .
The area is $\answer [given]{10}$ .

$\text {Area} = \int _1^3 (4x - 3) \ dx$

An anti-derivative of $4x-3$ is $2x^2 - 3x$

The Fundamental Theorem says: $\int _a^b f(x) \ dx = F(b) - F(a)$

Find the exact area under the curve $y = \sec ^2(x)$ from $x=0$ to $x= \pi /4$ .
Since the function $\cos (x)$ is non-negative on the interval $[0, \frac {\pi }{4}]$ , the exact area can be expressed as the definite integral $\int _0^{\pi /4} \sec ^2(x) \ dx.$ We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows: $\int _0^{\pi /4} \sec ^2(x) \ dx = \tan (x) \Big |_0^{\pi /4} = \tan (\tfrac {\pi }{4}) - \tan (0) = 1-0 =1.$

Find the exact area under the graph of $y = \csc ^2(x)$ from $x = \pi /4$ to $x = \pi /2$ .
The area is $\answer [given]{1}$ .

$\text {Area} = \int _{\pi /4}^{\pi /2} \csc ^2(x) \ dx$

An anti-derivative of $\csc ^2(x)$ is $-\cot (x)$

The Fundamental Theorem says: $\int _a^b f(x) \ dx = F(b) - F(a)$

$\cot (x) = \frac {\cos (x)}{\sin (x)}$

Find the exact area under the curve $y = \dfrac {1}{1+x^2}$ from $x=0$ to $x= 1$ .
Since the function $\frac {1}{1+x^2}$ is non-negative on the interval $[0, 1]$ , the exact area can be expressed as the definite integral $\int _0^{1} \frac {1}{1+x^2} \ dx.$ We can use the Fundamental Theorem of Calculus to compute the value of this definite integral as follows: $\int _0^{1} \frac {1}{1+x^2} \ dx = \tan ^{-1}(x) \Big |_0^1 = \tan ^{-1}(1)- \tan ^{-1}(0) = \tfrac {\pi }{4}-0 =\tfrac {\pi }{4}.$

Find the exact area under the graph of $y = 1/x$ from $x = 2$ to $x = 5$ .
The area is $\answer [given]{\pi /2}$ .

$\text {Area} = \int _0^1 \frac {1}{\sqrt {1-x^2}} \ dx$

An anti-derivative of $\frac {1}{\sqrt {1-x^2}}$ is $\sin ^{-1}(x)$

The Fundamental Theorem says: $\int _a^b f(x) \ dx = F(b) - F(a)$

$\sin ^{-1}(0) = 0$ and $\sin ^{-1}(1) = \frac {\pi }{2}$

We now consider the definite integral of negative functions. If $f(x) \leq 0$ for all $x$ in the interval $[a,b]$ , then $\sum _{i=1}^n f(x_i^*) \Delta x \leq 0,$ for any Riemann Sum of $f(x)$ on $[a, b]$ . Hence, the definite integral, $\int _a^b f(x) \ dx = \lim _{n \to \infty } \sum _{i=1}^n f(x_i^*) \Delta x \leq 0$ as well. How should we interpret the definite integral in this case? The graph of $y = f(x)$ will lie below the $x$ -axis, and the definite integral will equal $-1$ times the area of the region above the curve.

If $f(x)$ is continuous on the interval $[a, b]$ and if $f(x) \leq 0$ on $[a,b]$ , then $\int _a^b f(x) \ dx \leq 0,$

and

$\int _a^b f(x) \ dx = (-1) \cdot \text {area above the curve}.$

To put it another way, if $f(x) \leq 0$ on $[a,b]$ then $\text {area above the curve} = -\int _a^b f(x) \ dx.$

Find the exact area above the graph of $y = x^3$ from $x = -1$ to $x = 0$ .
Since the function $x^3$ is negative (or zero) on the interval $[-1, 0]$ , the exact area above the curve is the negative of the definite integral $\int _{-1}^0 x^3 \ dx.$ We can compute the value of this definite integral using the Fundamental Theorem of Calculus as follows: $\int _{-1}^0 x^3 \ dx = \frac {x^4}{4} \Big |_{-1}^0 = \frac {0^4}{4} - \frac {(-1)^4}{4} = -\frac 14.$

Hence the area of the region above the curve is $\frac 14$ .

Find the exact area above the graph of $y = x^3$ from $x = -2$ to $x = -1$ .

Area above the curve $= -\int _a^b f(x) \ dx$

$\int _{-2}^{-1} x^3 \ dx = \answer {-15/4},$ $\text {area above the curve} = \answer {15/4}.$

Find the exact area above the graph of $y = x^2 - 2$ from $x = 0$ to $x = 1$ .
Since the function $x^2 - 2 \leq 0$ on the interval $[0,1]$ , the exact area above the curve is the negative of the definite integral $\int _0^1 (x^2 -2) \ dx.$ We can compute the value of this definite integral using the Fundamental Theorem of Calculus as follows: $\int _0^1 (x^2 -2) \ dx = \left (\frac {x^3}{3}-2x\right ) \Bigg |_{0}^1 = \left (\frac {1^3}{3}-2(1)\right ) - \left (\frac {0^3}{3}- 2(0)\right ) = \frac 13 - 2 =-\frac 53.$

Hence the area of the region above the curve is $-\left (-\frac 53\right ) = \frac 53$ .

Find the exact area above the graph of $y = 1 - \sqrt x$ from $x = 1$ to $x = 4$ .

Area above the curve $= -\int _a^b f(x) \ dx$

$\int _1^4 1 - \sqrt x \ dx = \answer {-2/3},$ $\text {area above the curve} = \answer {2/3}.$

Use geometry to find the value of the definite integral $\displaystyle {\int _{-4}^{4}-\sqrt {16-x^2} \ dx}.$

$-\sqrt {16-x^2} \leq 0$ on the interval $[-4,4]$ .

The definite integral gives $(-1) \cdot$ area above the semi-circle

$\int _{-4}^{4}-\sqrt {16-x^2} \ dx = \answer {-8\pi }.$

Find the exact area above the graph of $y = -\sin (x)$ from $x = 0$ to $x = \pi$ .
Since the function $-\sin (x) \leq 0$ on the interval $[0, \pi ]$ , the exact area above the curve is the negative of the definite integral $\int _{0}^\pi -\sin (x) \ dx.$ We can compute the value of this definite integral using the Fundamental Theorem of Calculus as follows: $\int _0^\pi -\sin (x) \ dx = \cos (x) \Big |_{0}^\pi = \cos (\pi ) - \cos (0) = -1 -1 = -2.$

Hence the area of the region above the curve is $2$ .

Find the exact area above the graph of $y = \cos (x)$ from $x = \frac {\pi }{2}$ to $x = \frac {3\pi }{2}$ .

Area above the curve $= -\int _a^b f(x) \ dx$

$\int _{\pi /2}^{3\pi /2} \cos (x) \ dx = \answer {-2},$ $\text {area above the curve} = \answer {2}.$

Find the exact area above the graph of $y = \dfrac {1}{x}$ from $x = -3$ to $x = -1$ .

Area above the curve $= -\int _a^b f(x) \ dx$

$\int _{-3}^{-1} \frac {1}{x} \ dx = \answer {-\ln (3)},$ $\text {area above the curve} = \answer {\ln (3)}.$

Here are some detailed, lecture style videos on the definite integral: