Question

$\int_1^0 ((1-x^{25})\frac{20}{1} - (1-x^{20})\frac{25}{1}) dx$ (VII)

Answer 1

$\frac{1250}{273}$

Answer 2

The problem asks us to evaluate the definite integral $\int_1^0 ((1-x^{25})\frac{20}{1} - (1-x^{20})\frac{25}{1}) dx$.

First, simplify the integrand: $(1-x^{25})20 - (1-x^{20})25$ $= 20 - 20x^{25} - (25 - 25x^{20})$ $= 20 - 20x^{25} - 25 + 25x^{20}$ $= 25x^{20} - 20x^{25} - 5$

Now, substitute this simplified expression back into the integral: $I = \int_1^0 (25x^{20} - 20x^{25} - 5) dx$

We can integrate term by term using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$: $\int (25x^{20} - 20x^{25} - 5) dx = 25\frac{x^{20+1}}{20+1} - 20\frac{x^{25+1}}{25+1} - 5x$ $= 25\frac{x^{21}}{21} - 20\frac{x^{26}}{26} - 5x$ $= \frac{25}{21}x^{21} - \frac{10}{13}x^{26} - 5x$

Now, evaluate the definite integral using the limits from 1 to 0: $I = \left[ \frac{25}{21}x^{21} - \frac{10}{13}x^{26} - 5x \right]_1^0$ Apply the Fundamental Theorem of Calculus, which states $\int_a^b f(x) dx = F(b) - F(a)$: $I = \left( \frac{25}{21}(0)^{21} - \frac{10}{13}(0)^{26} - 5(0) \right) - \left( \frac{25}{21}(1)^{21} - \frac{10}{13}(1)^{26} - 5(1) \right)$ $I = (0 - 0 - 0) - \left( \frac{25}{21} - \frac{10}{13} - 5 \right)$ $I = - \left( \frac{25}{21} - \frac{10}{13} - 5 \right)$ $I = - \frac{25}{21} + \frac{10}{13} + 5$

To combine these fractions, find a common denominator for 21 and 13. Since 21 and 13 are coprime, the least common multiple (LCM) is $21 \times 13 = 273$. $I = \frac{-25 \times 13}{21 \times 13} + \frac{10 \times 21}{13 \times 21} + \frac{5 \times 273}{273}$ $I = \frac{-325}{273} + \frac{210}{273} + \frac{1365}{273}$ $I = \frac{-325 + 210 + 1365}{273}$ $I = \frac{-115 + 1365}{273}$ $I = \frac{1250}{273}$