The Fundamental Theorem of Calculus is a theorem that connects the two branches of calculus, differential and integral, into a single framework. We saw the computation of antiderivatives previously is the same process as integration; thus we know that differentiation and integration are inverse processes. The Fundamental Theorem of Calculus formalizes this connection. The theorem is given in two parts.

First Fundamental Theorem of Calculus¶

The first Fundamental Theorem of Calculus states:

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

$$ g(x) = \int_a^x f(t) \space dt \qquad a \leq x \leq b $$

is continuous on the interval $[a, b]$ and differentiable on $(a,b)$ and $g^\prime(x) = f(x)$.

Second Fundamental Theorem of Calculus¶

The second Fundamental Theorem of Calculus states:

If $f$ is continuous on the interval $[a, b]$ then:

$$ \int_a^b f(x) \space dx = F(b) - F(a) $$

Where $F$ is any antiderivative of $f$

Examples¶

In [2]:

from sympy import symbols, limit, diff, sin, cos, log, tan, sqrt, init_printing, plot, integrate
from mpmath import ln, e, pi

init_printing()
x = symbols('x')
y = symbols('y')

Example 1: Evaluate the integral: $\int_{-1}^2 (x^3 - 2x) \space dx$¶

Applying the second part of the Fundamental Theorem of Calculus, we take the antiderivative of the function and evaluate the integral.

$$ \int_{-1}^2 (x^3 - 2x) \space dx = \frac{1}{4} x^4 - x^2 \Bigg\rvert_{-1}^2 $$$$ = \frac{1}{4} (-1)^4 - (-1)^2 - \frac{1}{4} (2)^4 - (2)^2 = \frac{3}{4} $$

We can verify our answer using SymPy's integrate() function.

In [3]:

integrate(x ** 3 - 2 * x, (x, -1, 2))

Out[3]:

$$\frac{3}{4}$$

Example 2: Evaluate $\int_1^4 (5 - 2x + 3x^2) \space dx$¶

As in the previous example, we take advantage of the second part of the Fundamental Theorem of Calculus:

$$ \int_1^4 (5 - 2x + 3x^2) \space dx = 5x - x^2 + x^3 \Bigg\rvert_1^4 $$$$ = 5(4) - (4)^2 + (4)^3 - 5(1) - (1)^2 + (1)^3 = 63 $$

In [5]:

integrate(5 - 2 * x + 3 * x ** 2, (x, 1, 4))

Out[5]:

$$63$$

Example 3: Compute the integral $\int_0^1 x^{\frac{4}{5}} \space dx$¶

$$ \int_0^1 x^{\frac{4}{5}} \space dx = \frac{5}{9} x^{\frac{9}{5}} \Bigg\rvert_0^1 $$$$ = \frac{5}{9}(1)^\frac{9}{5} - \frac{5}{9}(0)^\frac{9}{5} = \frac{5}{9} $$

In [21]:

integrate(x ** (4/5), (x, 0, 1)) # Returned result will be in decimal form.

Out[21]:

$$0.555555555555556$$

Example 4: Determine the integral $\int_1^2 \frac{3}{x^4} \space dx$¶

Rewriting the integral as $\int_1^2 3x^{-4} \space dx$:

$$ \int_1^2 3x^{-4} \space dx = -x^{-3} = -\frac{1}{x^3} \Bigg\rvert_1^2 $$$$ = -\frac{1}{(2)^3} + \frac{1}{(1)^3} = -\frac{1}{8} + 1 = \frac{7}{8} $$

In [11]:

integrate(3 / x ** 4, (x, 1, 2))

Out[11]:

$$\frac{7}{8}$$

Example 5: Compute the integral $\int_0^2 x(2 + x^5) \space dx$¶

Start by factoring:

$$ \int_0^2 2x + x^6 \space dx = x^2 + \frac{1}{7} x^7 \Bigg\rvert_0^2 $$$$ = (2)^2 + \frac{1}{7} 2^7 - (0)^2 + \frac{1}{7} (0)^7 = 4 + \frac{128}{7} = \frac{28}{7} + \frac{128}{7} = \frac{156}{7} $$

In [20]:

integrate(x * (2 + x ** 5), (x, 0, 2))

Out[20]:

$$\frac{156}{7}$$

References¶

Fundamental theorem of calculus. (2017, December 2). In Wikipedia, The Free Encyclopedia. From https://en.wikipedia.org/w/index.php?title=Fundamental_theorem_of_calculus&oldid=813270221

Stewart, J. (2007). Essential calculus: Early transcendentals. Belmont, CA: Thomson Higher Education.

Weisstein, Eric W. "Fundamental Theorems of Calculus." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/FundamentalTheoremsofCalculus.html

The Fundamental Theorem of Calculus