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int_(0)^(2)2x((x^(2))/(2)+3)^(5)dx

$\int_{0}^{2} 2 x\left(\frac{x^{2}}{2}+3\right)^{5} d x$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\int_{0}^{2} 2 x\left(\frac{x^{2}}{2}+3\right)^{5} d x

Use Substitution: Let's use substitution. Let $u = \frac{x^2}{2} + 3$ . Then, $du = x dx$ . We need to adjust the $2x dx$ to fit $du$ , so we multiply $du$ by $2$ to get $2du = 2x dx$ .
Adjust for du: Now, we substitute into the integral: $\int_{0}^{2} 2x\left(\frac{x^2}{2} + 3\right)^5dx = \int_{u(0)}^{u(2)} u^5 \cdot 2du$ .
Substitute into Integral: We need to find the new limits of integration for $u$ . When $x = 0$ , $u = (0^2/2 + 3) = 3$ . When $x = 2$ , $u = (2^2/2 + 3) = 4 + 3 = 7$ . So, the new limits are from $3$ to $7$ .
Find New Limits for $u$ : Now we can integrate: $\int_{3}^{7} u^5 \cdot 2\,du = 2 \cdot \int_{3}^{7} u^5 \,du$ .
Integrate $u^5$ : Integrating $u^5$ gives us $(1/6)u^6$ . So, we have $2 \times (1/6)u^6$ from $3$ to $7$ .
Calculate Powers: Plugging in the limits, we get $2 \times \left(\frac{1}{6}\right) \times \left(7^6 - 3^6\right)$ .
Subtract Numbers: Calculating the powers, we get $2 \times \left(\frac{1}{6}\right) \times (117649 - 729)$ .
Multiply Through: Subtracting the two numbers gives us $2 \times \left(\frac{1}{6}\right) \times 116920$ .
Final Calculation: Multiplying through, we get $(\frac{2}{6}) \times 116920 = (\frac{1}{3}) \times 116920$ .
Final Calculation: Multiplying through, we get $(\frac{2}{6}) \times 116920 = (\frac{1}{3}) \times 116920$ . Finally, we calculate $(\frac{1}{3}) \times 116920$ to get $38973.33$ , but this is where I made a mistake. The correct calculation should be $(\frac{1}{3}) \times 116920 = 38973.333...$ , which simplifies to $38973 + \frac{1}{3}$ .

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