The Correct Answer and Explanation is:
The integral given is:
∫12x2e2x dx\int \frac{1}{2} x^2 e^{2x} \, dx
To evaluate this, we use integration by parts. The formula is:
∫u dv=uv−∫v du\int u \, dv = uv – \int v \, du
Choosing u=x2u = x^2 and dv=12e2x dxdv = \frac{1}{2} e^{2x} \, dx, we compute:
du=2x dx,v=14e2xdu = 2x \, dx, \quad v = \frac{1}{4} e^{2x}
Applying the formula:
∫12x2e2x dx=x2⋅14e2x−∫14e2x⋅2x dx\int \frac{1}{2} x^2 e^{2x} \, dx = x^2 \cdot \frac{1}{4} e^{2x} – \int \frac{1}{4} e^{2x} \cdot 2x \, dx
=14x2e2x−12∫xe2x dx= \frac{1}{4} x^2 e^{2x} – \frac{1}{2} \int x e^{2x} \, dx
We apply integration by parts again for ∫xe2x dx\int x e^{2x} \, dx, setting u=xu = x, dv=e2x dxdv = e^{2x} \, dx, yielding:
du=dx,v=12e2xdu = dx, \quad v = \frac{1}{2} e^{2x}
∫xe2x dx=x⋅12e2x−∫12e2x dx\int x e^{2x} \, dx = x \cdot \frac{1}{2} e^{2x} – \int \frac{1}{2} e^{2x} \, dx
=12xe2x−14e2x= \frac{1}{2} x e^{2x} – \frac{1}{4} e^{2x}
Substituting back:
∫12x2e2x dx=14x2e2x−14xe2x+18e2x+C\int \frac{1}{2} x^2 e^{2x} \, dx = \frac{1}{4} x^2 e^{2x} – \frac{1}{4} x e^{2x} + \frac{1}{8} e^{2x} + C
where CC is the constant of integration.
Explanation
Integration by parts is a method for evaluating integrals involving products of functions. The formula states that when integrating u dvu \, dv, the result involves the product uvuv minus the integral of v duv \, du. In this case, choosing u=x2u = x^2 and dvdv as the exponential term allowed us to simplify step by step.
After applying integration by parts the first time, the remaining integral still had an xx term, requiring another application of integration by parts. Breaking the problem into manageable parts is critical in such cases.
The choice of uu and dvdv should always prioritize reducing complexity with each step. Selecting x2x^2 allows differentiation to produce simpler terms, while choosing the exponential function for dvdv results in an integral that remains manageable.
After applying integration by parts twice, the final solution expresses the integral in terms of exponentials multiplied by polynomial terms. The presence of CC ensures that all indefinite integrals account for arbitrary constants.
This approach provides a systematic way to solve such problems, ensuring clarity and precision in each step.
