Evaluate The Integral. 64 2 3 X X Dx 1

Evaluating the Definite Integral (\displaystyle \int_{0}^{1} 64\cdot 2\cdot 3;x;x;dx) In this article we break down the computation of a simple polynomial definite integral step‑by‑step, explain the underlying calculus concepts, and show how to verify the result with alternative methods. The main keyword “evaluate the integral” appears naturally throughout the text to support SEO while keeping the discussion readable and engaging.

Introduction

When faced with an expression such as

[ \int_{0}^{1} 64;2;3;x;x;dx, ]

the first task is to interpret the notation correctly. The integral sign (\int) together with the differential (dx) and the limits (0) and (1) tells us we are dealing with a definite integral of a function over the interval ([0,1]). The integrand is the product of the constants (64), (2), and (3) multiplied by (x) twice, i.e., (x\cdot x = x^{2}).

Thus the integral to evaluate is

[ \int_{0}^{1} \bigl(64\cdot 2\cdot 3\bigr),x^{2};dx = \int_{0}^{1} 384,x^{2};dx . ]

The goal of this article is to evaluate the integral completely, from setting up the problem to presenting the final numeric answer, while also discussing why each step works. By the end, readers will have a clear template for tackling similar polynomial definite integrals.

Step‑by‑Step Evaluation

1. Simplify the Constant Coefficient

Multiply the numerical constants that appear in front of the variable part:

[64 \times 2 \times 3 = (64 \times 2) \times 3 = 128 \times 3 = 384 . ]

Hence the integrand becomes (384x^{2}).

Key point: Constants can be factored out of the integral (or multiplied first) because integration is a linear operation.

2. Apply the Power Rule for Integration

Recall the power rule for antiderivatives:

[ \int x^{n},dx = \frac{x^{n+1}}{n+1} + C \qquad (n\neq -1). ]

For our integrand (384x^{2}), the exponent (n = 2). Applying the rule:

[ \int 384x^{2},dx = 384 \int x^{2},dx = 384 \left(\frac{x^{2+1}}{2+1}\right) + C = 384 \left(\frac{x^{3}}{3}\right) + C = 128x^{3} + C . ]

Thus the indefinite integral (antiderivative) is (F(x)=128x^{3}+C).

3. Evaluate the Definite Integral Using the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (Part 2) states:

[\int_{a}^{b} f(x),dx = F(b)-F(a), ]

where (F) is any antiderivative of (f). Using (F(x)=128x^{3}) (the constant (C) cancels out), we compute:

[ \begin{aligned} \int_{0}^{1} 384x^{2},dx &= \bigl[128x^{3}\bigr]_{0}^{1} \ &= 128(1)^{3} - 128(0)^{3} \ &= 128 - 0 \ &= 128 . \end{aligned} ]

Therefore, the value of the definite integral is 128.

4. Alternative Verification (Riemann Sum Insight)

To reinforce confidence, we can view the integral as the limit of a Riemann sum. Partition ([0,1]) into (n) equal subintervals of width (\Delta x = \frac{1}{n}). Choose the right‑hand endpoint (x_i^{*}= \frac{i}{n}) as sample point. Then:

[ \begin{aligned} \sum_{i=1}^{n} f(x_i^{*})\Delta x &= \sum_{i=1}^{n} 384\left(\frac{i}{n}\right)^{2}\left(\frac{1}{n}\right) \ &= \frac{384}{n^{3}} \sum_{i=1}^{n} i^{2}. \end{aligned} ]

Using the known formula (\displaystyle \sum_{i=1}^{n} i^{2}= \frac{n(n+1)(2n+1)}{6}),

[\frac{384}{n^{3}} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{384}{6}\cdot \frac{(n+1)(2n+1)}{n^{2}} = 64 \cdot \frac{(n+1)(2n+1)}{n^{2}} . ]

As (n\to\infty), (\frac{(n+1)(2n+1)}{n^{2}} \to 2). Hence the limit equals (64 \times 2 = 128), matching our earlier result.

Conceptual Discussion

Why Constants Factor Out Integration is a linear operator: for any constants (a) and (b),

[ \int \bigl[a f(x) + b g(x)\bigr]dx = a\int f(x)dx + b\int g(x)dx . ]

In our case, the integrand is a single term (384x^{2}); we can treat (384) as the constant (a) and (f(x)=x^{2}). Factoring it out simplifies the power rule application.

The Role of the Power Rule

The power rule emerges from reversing the power rule for differentiation. Since (\frac{d}{dx}\bigl(x^{n+1}\bigr) = (n

5. Conclusion

We have successfully evaluated the definite integral of (384x^2) from 0 to 1, demonstrating that the value is 128. This was achieved by applying the power rule for integration, utilizing the Fundamental Theorem of Calculus, and confirming the result through a Riemann sum approach. The key to this solution lies in recognizing the linearity of integration and leveraging the power rule to find the antiderivative. The Fundamental Theorem of Calculus then allows us to calculate the definite integral by evaluating the antiderivative at the upper and lower limits of integration and subtracting the values. Finally, the Riemann sum provides a powerful geometric and algebraic verification, solidifying our understanding of the integral's value. This example highlights the importance of understanding the fundamental concepts of integration and how they connect to calculus's core principles. The ability to apply these techniques allows us to solve a wide range of problems involving areas, volumes, and other quantities that can be expressed as integrals.

Building on the Riemann‑sum verification, one can also view the integral through the lens of the Beta function. Recall that for (p,q>-1),

[ \int_{0}^{1} x^{p}(1-x)^{q},dx = B(p+1,q+1)=\frac{\Gamma(p+1)\Gamma(q+1)}{\Gamma(p+q+2)}. ]

Setting (q=0) reduces the Beta integral to (\int_{0}^{1}x^{p},dx = 1/(p+1)). With (p=2) we recover the familiar factor (1/3), and multiplying by the constant 384 yields (384/3 = 128). This perspective highlights how definite integrals of monomials are special cases of a broader family of integrals that appear frequently in probability theory (e.g., moments of the Beta distribution) and in physics when normalizing wavefunctions.

Another useful check comes from a simple substitution. Let (u = x^{3}); then (du = 3x^{2}dx) and the limits transform as (x=0\Rightarrow u=0) and (x=1\Rightarrow u=1). The integral becomes

[ \int_{0}^{1}384x^{2},dx = 384\int_{0}^{1}\frac{1}{3},du = 128\int_{0}^{1}du =128, ]

confirming the result while illustrating how a change of variables can reduce a polynomial integrand to a constant.

These alternative viewpoints reinforce the robustness of the answer and showcase the interconnectedness of integration techniques—linearity, the power rule, the Fundamental Theorem of Calculus, Riemann sums, special functions, and substitution—all converging on the same numerical value.

Conclusion

Through direct application of the power rule, verification via Riemann sums, connection to the Beta function, and a straightforward substitution, we have consistently found that

[ \int_{0}^{1}384x^{2},dx = 128. ]

The agreement among these diverse methods not only confirms the correctness of the computation but also deepens our appreciation for the coherence of calculus as a unified framework. Mastery of such complementary strategies equips us to tackle more complex integrals with confidence, whether they arise in pure mathematics, applied sciences, or engineering contexts.

While the integral (\int_{0}^{1} 384x^{2},dx) is elementary, the multiplicity of verification routes—from the mechanical power rule to the abstract Beta function—serves as a microcosm of mathematical problem-solving. In practice, integrals rarely submit to a single, obvious technique; instead, they often demand a synthesis of substitutions, series expansions, or transformations into known special functions. The discipline of cross-validating results through independent methods is not merely academic—it is a safeguard against error in fields where integrals model real-world systems, from calculating work in non-uniform force fields to evaluating probabilities in Bayesian statistics.

Moreover, the very act of recognizing a polynomial integral as a Beta function instance illustrates a deeper skill: pattern recognition across domains. The Beta function’s appearance here hints at its ubiquity in normalizing distributions, solving differential equations, and even in evaluating certain trigonometric integrals via Euler’s formula. Such connections transform isolated computations into nodes within a vast network of ideas, where a technique borrowed from probability can simplify a geometric area problem, or a substitution from physics can demystify an algebraic expression.

Ultimately, this exploration underscores that the power of calculus lies not just in its ability to compute but in its capacity to unify. The consistent result—128—emerges from a confluence of perspectives, each illuminating a different facet of the integral’s structure. As we advance to more intricate integrands, whether involving transcendental functions, multiple variables, or complex paths, the habit of seeking multiple avenues of approach remains indispensable. It is through such layered understanding that we move from routine calculation to genuine insight, appreciating calculus not as a collection of formulas but as a coherent language describing change, accumulation, and connection.