Fundamental Theorem of Calculus on Integrals
We have learned about definite integrals in the previous section. To determine the value of a definite integral using Riemann sums requires complicated steps. Newton and Leibniz discovered an easier way to determine the value of definite integrals. This method is known as the Fundamental Theorem of Calculus (FTC).
In the following discussion, we will learn about the fundamental theorem of calculus. The fundamental theorem of calculus consists of FTC I and FTC II. This theorem is widely used in applied problems, for example finding the area of a region bounded by a curve or determining the volume of a solid of revolution, which certainly involves calculations using integrals.
Generally, FTC I states the inverse relationship of integration - if there is a function in integral form, then to remove the integral we use FTC I. Meanwhile, FTC II is about how to calculate definite integral forms.
Fundamental Theorem of Calculus I and II
If \( f \) is continuous on \( [a, b] \) and \( x \) is any point in \( (a, b) \), then:
Note: \( a \) is the lower limit of integration which is a constant and does not affect the result.
If \( f \) is continuous on \( [a, b] \) and \( F \) is an antiderivative of \( f \) on \( [a, b] \), then:
The form \( F(b) - F(a) \) can be written as \( [F(x)]_a^b \), so FTC II can be written as:
Example Problems
Find the results of:
a) \(\frac{d}{dx} \int_0^x (3t^2 - t + 6) dt\)
b) \(\frac{d}{dx} \int_{-5}^{x} \left(\frac{1}{3}t^2 + 1\right) dt\)
Solution: We use the Fundamental Theorem of Calculus I:
a) \(\frac{d}{dx} \int_0^x (3t^2 - t + 6) dt\)
This means the function \( f(t) = 3t^2 - t + 6 \), so \( f(x) = 3x^2 - x + 6 \)
Therefore, \(\frac{d}{dx} \int_0^x (3t^2 - t + 6) dt = 3x^2 - x + 6\)
b) \(\frac{d}{dx} \int_{-5}^{x} \left(\frac{1}{3}t^2 + 1\right) dt\)
This means the function \( f(t) = \frac{1}{3}t^2 + 1 \), so \( f(x) = \frac{1}{3}x^2 + 1 \)
Therefore, \(\frac{d}{dx} \int_{-5}^{x} \left(\frac{1}{3}t^2 + 1\right) dt = \frac{1}{3}x^2 + 1\)
Find the results of the following integrals:
a) \(\int_{-1}^{1} (3x^2 + 2x - 1) dx\)
b) \(\int_{-2}^{1} (x + 5) dx\)
Solution: We use the Fundamental Theorem of Calculus II:
a) \(\int_{-1}^{1} (3x^2 + 2x - 1) dx\)
General integration formula: \(\int a x^n dx = \frac{a}{n+1} x^{n+1} + C\)
\(\int_{-1}^{1} (3x^2 + 2x - 1) dx = \left[\frac{3}{3}x^3 + \frac{2}{2}x^2 - x\right]_{-1}^{1}\)
\(= \left[x^3 + x^2 - x\right]_{-1}^{1}\)
\(= \left[(1)^3 + (1)^2 - (1)\right] - \left[(-1)^3 + (-1)^2 - (-1)\right]\)
\(= \left[1 + 1 - 1\right] - \left[-1 + 1 + 1\right]\)
\(= [1] - [1] = 0\)
Wait, let's recalculate carefully:
\(F(x) = x^3 + x^2 - x\)
\(F(1) = 1^3 + 1^2 - 1 = 1 + 1 - 1 = 1\)
\(F(-1) = (-1)^3 + (-1)^2 - (-1) = -1 + 1 + 1 = 1\)
\(\int_{-1}^{1} (3x^2 + 2x - 1) dx = F(1) - F(-1) = 1 - 1 = 0\)
Result: \(\int_{-1}^{1} (3x^2 + 2x - 1) dx = 0\)
b) \(\int_{-2}^{1} (x + 5) dx\)
\(\int_{-2}^{1} (x + 5) dx = \left[\frac{1}{2}x^2 + 5x\right]_{-2}^{1}\)
\(= \left[\frac{1}{2}(1)^2 + 5(1)\right] - \left[\frac{1}{2}(-2)^2 + 5(-2)\right]\)
\(= \left[\frac{1}{2} + 5\right] - \left[\frac{1}{2}(4) - 10\right]\)
\(= \left[5.5\right] - \left[2 - 10\right]\)
\(= 5.5 - (-8) = 5.5 + 8 = 13.5\)
Result: \(\int_{-2}^{1} (x + 5) dx = 13.5\) or \(13\frac{1}{2}\)
If given the function \( f(x) = \int_{-1}^{x} (t^4 + t - 1) dt \), then the value of \( f'(1) = \ldots \), where \( f'(x) \) is the first derivative of \( f(x) \).
Solution:
From the form \( f(x) = \int_{-1}^{x} (t^4 + t - 1) dt \), we differentiate both sides using FTC I:
\(f(x) = \int_{-1}^{x} (t^4 + t - 1) dt \) (differentiate both sides)
\(\frac{d}{dx} f(x) = \frac{d}{dx} \int_{-1}^{x} (t^4 + t - 1) dt\)
\(f'(x) = x^4 + x - 1\) (by FTC I)
Thus, \( f'(1) = 1^4 + 1 - 1 = 1 + 1 - 1 = 1 \)
Result: \( f'(1) = 1 \)
Calculate the limit: \(\lim_{x \to 0} \frac{\int_{0}^{x} \sin(t^3) dt}{x^4}\)
Solution:
Step 1: Direct substitution \( x = 0 \):
\(\lim_{x \to 0} \frac{\int_{0}^{x} \sin(t^3) dt}{x^4} = \frac{\int_{0}^{0} \sin(t^3) dt}{0^4} = \frac{0}{0}\)
Since the result is \( \frac{0}{0} \) (indeterminate form), we process further using L'Hospital's Rule (derivative).
Step 2: Determine derivatives of numerator and denominator:
Numerator: \( y = \int_{0}^{x} \sin(t^3) dt \Rightarrow y' = \frac{d}{dx} \int_{0}^{x} \sin(t^3) dt = \sin(x^3) \) (by FTC I)
Denominator: \( y = x^4 \Rightarrow y' = 4x^3 \)
Step 3: Apply L'Hospital's Rule:
\(\lim_{x \to 0} \frac{\int_{0}^{x} \sin(t^3) dt}{x^4} = \lim_{x \to 0} \frac{\sin(x^3)}{4x^3}\)
Substitute \( x = 0 \): \( \frac{\sin(0)}{0} = \frac{0}{0} \) (still indeterminate)
Step 4: Apply L'Hospital's Rule again:
Numerator: \( y = \sin(x^3) \Rightarrow y' = 3x^2 \cos(x^3) \)
Denominator: \( y = 4x^3 \Rightarrow y' = 12x^2 \)
\(\lim_{x \to 0} \frac{\sin(x^3)}{4x^3} = \lim_{x \to 0} \frac{3x^2 \cos(x^3)}{12x^2} = \lim_{x \to 0} \frac{\cos(x^3)}{4} = \frac{\cos(0)}{4} = \frac{1}{4}\)
Result: \(\lim_{x \to 0} \frac{\int_{0}^{x} \sin(t^3) dt}{x^4} = \frac{1}{4}\)
Key Points to Remember
The derivative of an integral with a variable upper limit gives back the original function evaluated at that variable:
This shows that differentiation and integration are inverse operations.
To evaluate a definite integral, find an antiderivative \( F(x) \) of \( f(x) \), then calculate:
This is much simpler than using Riemann sums!
• Finding areas under curves
• Calculating volumes of solids of revolution
• Solving problems in physics (displacement, work, etc.)
• Evaluating limits involving integrals (using L'Hospital's Rule with FTC I)
The Fundamental Theorem of Calculus was discovered independently by Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century. This theorem forms the foundation of calculus and connects the two main branches: differentiation and integration.