Step 1: Understanding the Concept:
A powerful property of definite integrals is:
\[ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \]
This substitution is particularly useful when the integrand contains a factor like \((a - x)^n\).
By applying this property, the complex power moves to the simpler variable \(x\), making it much easier to distribute and integrate term by term without performing a binomial expansion of a high power.
Step 2: Key Formula or Approach:
Let \(I = \int_{0}^{2} x^2(2 - x)^5 dx\).
We substitute \(x\) with \((2 - x)\).
Step 3: Detailed Explanation:
Applying the property:
\[ I = \int_{0}^{2} (2 - x)^2 (2 - (2 - x))^5 dx \]
\[ I = \int_{0}^{2} (2 - x)^2 x^5 dx \]
Expanding the square term \((2 - x)^2 = 4 - 4x + x^2\):
\[ I = \int_{0}^{2} (4 - 4x + x^2) x^5 dx \]
Distribute \(x^5\) into the parentheses:
\[ I = \int_{0}^{2} (4x^5 - 4x^6 + x^7) dx \]
Integrating term by term:
\[ I = \left[ 4\frac{x^6}{6} - 4\frac{x^7}{7} + \frac{x^8}{8} \right]_{0}^{2} \]
Simplify the fractions:
\[ I = \left[ \frac{2}{3}x^6 - \frac{4}{7}x^7 + \frac{1}{8}x^8 \right]_{0}^{2} \]
Substitute the upper limit \(x = 2\):
\[ I = \frac{2}{3}(2^6) - \frac{4}{7}(2^7) + \frac{1}{8}(2^8) \]
\[ I = \frac{128}{3} - \frac{512}{7} + \frac{256}{8} \]
\[ I = \frac{128}{3} - \frac{512}{7} + 32 \]
To combine these, use the common denominator \(21\):
\[ I = \frac{128 \times 7 - 512 \times 3 + 32 \times 21}{21} \]
\[ I = \frac{896 - 1536 + 672}{21} \]
\[ I = \frac{1568 - 1536}{21} = \frac{32}{21} \]
Step 4: Final Answer:
The result of the integral is \(\frac{32}{21}\).