Integral of 1 × x² · 2 × x²: A Step‑by‑Step Guide
When faced with an expression like “1 x 2 2x 2 integral”, the first task is to rewrite it in standard mathematical notation. The most natural interpretation is the product of two monomials:
[ (1\cdot x^{2})\times(2\cdot x^{2}) ;=; 2x^{4}. ]
Thus the problem reduces to evaluating the indefinite integral
[ \int 2x^{4},dx . ]
Although the integrand looks simple, understanding why the power rule works and how to apply it reliably builds a foundation for tackling far more complicated integrals. The following sections walk through the concept, the mechanics, verification, practical uses, and common pitfalls—all in clear, accessible language.
Introduction: Why This Integral Matters
Integrals are the inverse operation of differentiation and appear everywhere—from physics (calculating work, energy, and momentum) to economics (finding consumer surplus) and engineering (determining areas under curves). Even a modest integral like (\int 2x^{4}dx) illustrates the core idea: increase the exponent by one and divide by the new exponent. Mastering this rule for monomials enables students to decompose polynomials, rational functions, and eventually trigonometric or exponential expressions into manageable pieces.
The phrase “1 x 2 2x 2 integral” may initially confuse readers because of missing caret symbols (^). By clarifying the intended meaning—(1\cdot x^{2}) multiplied by (2\cdot x^{2})—we set the stage for a precise calculation.
Understanding the Integrand
Breaking Down the Expression
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Identify each factor
- First factor: (1 \times x^{2}) → simply (x^{2}).
- Second factor: (2 \times x^{2}) → (2x^{2}).
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Multiply the factors
[ x^{2}\cdot 2x^{2}=2x^{2+2}=2x^{4}. ]
The product rule for exponents states that when multiplying like bases, you add the exponents: (x^{a}\cdot x^{b}=x^{a+b}). The numerical coefficients multiply as ordinary numbers (1 × 2 = 2) And that's really what it comes down to..
Visualizing the Function
The integrand (f(x)=2x^{4}) is an even, monotonic function for (x\ge0) and symmetric about the y‑axis. Its graph rises steeply as (|x|) grows, which influences the shape of its antiderivative (a quintic polynomial).
Step‑by‑Step Integration
Apply the Power Rule
The power rule for integration states:
[ \int x^{n},dx = \frac{x^{n+1}}{n+1}+C,\qquad n\neq -1. ]
Because our integrand includes a constant coefficient, we can factor it out:
[ \int 2x^{4},dx = 2\int x^{4},dx. ]
Now apply the rule with (n=4):
[ \int x^{4},dx = \frac{x^{5}}{5}+C_{1}. ]
Multiplying back the coefficient:
[ 2\left(\frac{x^{5}}{5}+C_{1}\right)=\frac{2}{5}x^{5}+C, ]
where (C=2C_{1}) absorbs the arbitrary constant Most people skip this — try not to..
Final Result
[ \boxed{\displaystyle \int 2x^{4},dx = \frac{2}{5}x^{5}+C }. ]
If a definite integral over an interval ([a,b]) is required, evaluate the antiderivative at the bounds:
[ \int_{a}^{b}2x^{4},dx = \left[\frac{2}{5}x^{5}\right]_{a}^{b} = \frac{2}{5}\bigl(b^{5}-a^{5}\bigr). ]
Verification by Differentiation
A quick way to confirm the antiderivative is to differentiate it:
[ \frac{d}{dx}\left(\frac{2}{5}x^{5}+C\right)=\frac{2}{5}\cdot5x^{4}=2x^{4}, ]
which returns the original integrand. This consistency check is a habit worth cultivating; it catches sign errors, missed coefficients, or incorrect exponent adjustments It's one of those things that adds up..
General Rule for Products of Monomials
The example above illustrates a broader principle:
When integrating a product of monomials, first combine like bases by adding exponents, multiply the numerical coefficients, then apply the power rule.
Symbolically, for constants (a,b) and exponents (p,q):
[ \int (a x^{p})(b x^{q}),dx = ab\int x^{p+q},dx = \frac{ab}{p+q+1}x^{p+q+1}+C,\quad p+q\neq -1. ]
If the sum of exponents equals (-1), the integral becomes a logarithmic form (\int x^{-1}dx=\ln|x|+C). Recognizing this special case prevents division‑by‑zero mistakes Easy to understand, harder to ignore..
Applications in Real‑World Contexts
Physics: Work Done by a Variable Force
Suppose a force (F(x)=2x^{4}) newtons acts on an object moving along the x‑axis. The work done from (x=a) to (x=b) is:
[ W=\int_{a}^{b}F(x),dx=\frac{2}{5}\bigl(b^{5}-a^{5}\bigr)\ \text{joules}. ]
Economics: Consumer Surplus
If the marginal willingness to pay for a good is given by (MWTP(x)=2x^{4}) dollars per unit, the total willingness to pay for consuming up to (x=b) units (starting from zero) is the same integral, representing the area under the demand curve.
Engineering: Moment of Inertia
For a thin rod with linear density varying as (\lambda(x)=2x^{4
