## The Fundamental Theorem of Calculus

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