讲解一下图中题目---**Question Stem:**
Evaluate the indefinite integral:
∫xe^(x²) dx
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Let's evaluate the indefinite integral of x times e to the power of x squared. This is an indefinite integral that requires special integration techniques. We can see that this integral involves a composite function, e to the x squared, multiplied by x. This structure suggests that we should use the substitution method. The key components are: the exponential function, the quadratic exponent x squared, and the linear factor x.
Now we'll use u-substitution to solve this integral. The key is choosing the right substitution. We let u equal x squared. Taking the derivative, we get du equals 2x dx, which means x dx equals one-half du. This transforms our original integral from x times e to the x squared dx into e to the u times one-half du. Notice how the x squared becomes u, and the x dx becomes one-half du. This color coding shows the correspondence between the original terms and the substituted terms.
Now we perform the actual integration. We have the simplified integral: one-half times the integral of e to the u du. We apply the basic integration rule for exponential functions: the integral of e to the u du equals e to the u plus C. This gives us one-half times e to the u plus C, which simplifies to one-half e to the u plus C. This step emphasizes the fundamental property that the exponential function integrates to itself.
Now we perform the back-substitution to convert our result back to the original variable. We have one-half e to the u plus C. Recalling that u equals x squared, we substitute back to get one-half e to the x squared plus C. This is our final answer. Let's verify this result by taking the derivative. The derivative of one-half e to the x squared plus C equals one-half times e to the x squared times 2x, using the chain rule. This simplifies to x times e to the x squared, which is exactly our original integrand. This confirms our solution is correct.
Let's analyze why this substitution method works and how to recognize similar problems. The key reason u equals x squared works is that the derivative of u, which is 2x, appears as a factor in our original integrand. This follows a general pattern: when integrating f prime of x times g of f of x, we use the substitution u equals f of x. Here are similar examples: the integral of x sine of x squared, x cosine of x squared, and x times x squared plus one to the fifth power. To recognize when substitution is appropriate, look for composite functions, check if the derivative of the inner function appears as a factor, and consider u-substitution whenever the chain rule is involved in differentiation.