Evaluating the Definite Integral (\displaystyle \int_{0}^{1} \frac{3}{7}x , dx)
The integral (\displaystyle \int_{0}^{1} \frac{3}{7}x , dx) is a simple linear integral that illustrates the core ideas of the Fundamental Theorem of Calculus. By mastering this example, you’ll gain confidence in handling more complex integrals, whether you’re studying calculus, engineering, physics, or any field that relies on area, accumulation, or average values.
Introduction
A definite integral represents the net signed area between a function and the x‑axis over a closed interval. In our case, the function is the linear function (\frac{3}{7}x), and the interval is ([0,1]). The calculation involves two key steps:
- Find an antiderivative (also called an indefinite integral) of the integrand.
- Apply the Fundamental Theorem of Calculus (FTC) by evaluating the antiderivative at the upper and lower limits and subtracting.
Although the integrand is straightforward, this example is an excellent opportunity to reinforce the process, avoid common pitfalls, and understand the geometric meaning behind the numbers Easy to understand, harder to ignore. Which is the point..
Step‑by‑Step Solution
1. Identify the Integrand and Limits
[ f(x) = \frac{3}{7}x,\qquad a = 0,\qquad b = 1 ]
2. Find an Antiderivative
The antiderivative (F(x)) satisfies (F'(x) = f(x)).
For a linear function (kx), the antiderivative is (\frac{k}{2}x^{2}).
Thus:
[
F(x) = \frac{3}{7}\cdot\frac{x^{2}}{2} = \frac{3}{14}x^{2}
]
3. Apply the Fundamental Theorem of Calculus
[ \int_{0}^{1} \frac{3}{7}x , dx = F(1) - F(0) ] Compute each term: [ F(1) = \frac{3}{14}(1)^{2} = \frac{3}{14} ] [ F(0) = \frac{3}{14}(0)^{2} = 0 ] Subtract: [ \frac{3}{14} - 0 = \boxed{\frac{3}{14}} ]
Geometric Interpretation
The graph of (y = \frac{3}{7}x) is a straight line passing through the origin with slope (\frac{3}{7}). Between (x = 0) and (x = 1), the area under the line and above the x‑axis forms a right triangle. The base is 1 unit long, and the height at (x = 1) is (\frac{3}{7}) The details matter here..
[ \text{Area} = \frac{1}{2}\times 1 \times \frac{3}{7} = \frac{3}{14} ]
This geometric calculation matches the analytical result, confirming the integral’s interpretation as an area Simple, but easy to overlook..
Common Mistakes to Avoid
| Mistake | Why It Happens | Correct Approach |
|---|---|---|
| Dropping the constant factor | Forgetting that constants pull out of the integral. | |
| Incorrect exponent after integration | Misapplying the power rule. Think about it: | |
| Reversing limits | Mixing up the order of subtraction. In practice, | (\int x,dx = \frac{x^2}{2}), not (\frac{x^3}{3}). |
| Neglecting the sign of the area | Assuming all areas are positive. On top of that, | Always evaluate (F(b)-F(a)). |
Variations and Extensions
| Variation | Result | Interpretation |
|---|---|---|
| (\displaystyle \int_{0}^{2} \frac{3}{7}x , dx) | (\frac{3}{14}\times 4 = \frac{6}{7}) | Triangle with base 2, height (\frac{3}{7}\times 2 = \frac{6}{7}). That's why |
| (\displaystyle \int_{-1}^{1} \frac{3}{7}x , dx) | (0) | Symmetric positive and negative areas cancel. |
| (\displaystyle \int_{0}^{1} \frac{3}{7}x^{2} , dx) | (\frac{3}{28}) | Area under a parabola, smaller than the linear case. |
These examples illustrate how changing limits or the power of (x) directly affects the area, reinforcing the linearity property of integrals.
Frequently Asked Questions (FAQ)
Q1: Can I integrate (\frac{3}{7}x) by treating it as (3x/7) or ((3/7)x)?
A1: Both notations are equivalent. The constant (\frac{3}{7}) can be factored out of the integral: (\frac{3}{7}\int x,dx).
Q2: Why is the antiderivative (\frac{3}{14}x^{2}) and not (\frac{3}{7}x^{2})?
A2: Integrating (x) yields (\frac{x^{2}}{2}). Multiplying by the constant (\frac{3}{7}) gives (\frac{3}{7}\times\frac{x^{2}}{2} = \frac{3}{14}x^{2}).
Q3: Does the definite integral always represent area?
A3: It represents net signed area. If the function dips below the x‑axis, the integral subtracts that portion. In this example the function stays above the axis, so the integral equals the geometric area.
Q4: How would I solve this integral using substitution?
A4: Substitution isn’t necessary for a linear function, but you could set (u = x), (du = dx). The integral becomes (\frac{3}{7}\int_{0}^{1} u,du), which evaluates the same way.
Conclusion
Evaluating (\displaystyle \int_{0}^{1} \frac{3}{7}x , dx) is a textbook demonstration of the Fundamental Theorem of Calculus. Think about it: by following the systematic steps—identifying the integrand, finding an antiderivative, and applying the limits—you arrive at the clear result (\frac{3}{14}). The geometric perspective, confirming the area of a right triangle, reinforces the connection between algebraic manipulation and visual intuition. Mastery of this simple example builds a solid foundation for tackling a wide range of integrals in mathematics, science, and engineering.
