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Sagot :
To solve the integral [tex]\(\int x e^{-2 x} \, dx\)[/tex], we can use integration by parts. Remember that integration by parts is given by the formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
First, we need to choose [tex]\(u\)[/tex] and [tex]\(dv\)[/tex]:
- Let [tex]\(u = x\)[/tex]. Then, its derivative [tex]\(du = dx\)[/tex].
- Let [tex]\(dv = e^{-2x} \, dx\)[/tex]. We need to find [tex]\(v\)[/tex] by integrating [tex]\(dv\)[/tex]. The integral of [tex]\(e^{-2x} \, dx\)[/tex] is [tex]\(\int e^{-2x} \, dx\)[/tex].
To integrate [tex]\(e^{-2x}\)[/tex], we perform a simple u-substitution. Let [tex]\(w = -2x\)[/tex]. Then, [tex]\(dw = -2 \, dx\)[/tex] or [tex]\(dx = -\frac{1}{2} \, dw\)[/tex]. Hence, we have:
[tex]\[ \int e^{-2x} \, dx = \int e^{w} \left(-\frac{1}{2} \, dw\right) = -\frac{1}{2} \int e^{w} \, dw = -\frac{1}{2} e^{w} = -\frac{1}{2} e^{-2x} \][/tex]
Thus, [tex]\(v = -\frac{1}{2} e^{-2x}\)[/tex].
Now, apply the integration by parts formula:
[tex]\[ \int x e^{-2 x} \, dx = uv - \int v \, du \][/tex]
Substitute [tex]\(u = x\)[/tex], [tex]\(du = dx\)[/tex], [tex]\(v = -\frac{1}{2} e^{-2x}\)[/tex], and [tex]\(dv = e^{-2x} \, dx\)[/tex]:
[tex]\[ \int x e^{-2x} \, dx = x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) dx \][/tex]
Simplify the expression:
[tex]\[ = -\frac{1}{2} x e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx \][/tex]
We have already found that [tex]\(\int e^{-2x} \, dx = -\frac{1}{2} e^{-2x}\)[/tex]. Substitute this result back into the expression:
[tex]\[ = -\frac{1}{2} x e^{-2x} + \frac{1}{2} \left(-\frac{1}{2} e^{-2x}\right) \][/tex]
Simplify further:
[tex]\[ = -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} \][/tex]
Factor out [tex]\(e^{-2x}\)[/tex] from the expression:
[tex]\[ = \left(-\frac{1}{2} x - \frac{1}{4}\right) e^{-2x} \][/tex]
For simplicity, we can combine the fractions:
[tex]\[ = \left(\frac{-2x - 1}{4}\right) e^{-2x} \][/tex]
Therefore, the solution to the integral [tex]\(\int x e^{-2x} \, dx\)[/tex] is:
[tex]\[ \boxed{\frac{(-2x - 1)}{4} e^{-2x} + C} \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
First, we need to choose [tex]\(u\)[/tex] and [tex]\(dv\)[/tex]:
- Let [tex]\(u = x\)[/tex]. Then, its derivative [tex]\(du = dx\)[/tex].
- Let [tex]\(dv = e^{-2x} \, dx\)[/tex]. We need to find [tex]\(v\)[/tex] by integrating [tex]\(dv\)[/tex]. The integral of [tex]\(e^{-2x} \, dx\)[/tex] is [tex]\(\int e^{-2x} \, dx\)[/tex].
To integrate [tex]\(e^{-2x}\)[/tex], we perform a simple u-substitution. Let [tex]\(w = -2x\)[/tex]. Then, [tex]\(dw = -2 \, dx\)[/tex] or [tex]\(dx = -\frac{1}{2} \, dw\)[/tex]. Hence, we have:
[tex]\[ \int e^{-2x} \, dx = \int e^{w} \left(-\frac{1}{2} \, dw\right) = -\frac{1}{2} \int e^{w} \, dw = -\frac{1}{2} e^{w} = -\frac{1}{2} e^{-2x} \][/tex]
Thus, [tex]\(v = -\frac{1}{2} e^{-2x}\)[/tex].
Now, apply the integration by parts formula:
[tex]\[ \int x e^{-2 x} \, dx = uv - \int v \, du \][/tex]
Substitute [tex]\(u = x\)[/tex], [tex]\(du = dx\)[/tex], [tex]\(v = -\frac{1}{2} e^{-2x}\)[/tex], and [tex]\(dv = e^{-2x} \, dx\)[/tex]:
[tex]\[ \int x e^{-2x} \, dx = x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) dx \][/tex]
Simplify the expression:
[tex]\[ = -\frac{1}{2} x e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx \][/tex]
We have already found that [tex]\(\int e^{-2x} \, dx = -\frac{1}{2} e^{-2x}\)[/tex]. Substitute this result back into the expression:
[tex]\[ = -\frac{1}{2} x e^{-2x} + \frac{1}{2} \left(-\frac{1}{2} e^{-2x}\right) \][/tex]
Simplify further:
[tex]\[ = -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} \][/tex]
Factor out [tex]\(e^{-2x}\)[/tex] from the expression:
[tex]\[ = \left(-\frac{1}{2} x - \frac{1}{4}\right) e^{-2x} \][/tex]
For simplicity, we can combine the fractions:
[tex]\[ = \left(\frac{-2x - 1}{4}\right) e^{-2x} \][/tex]
Therefore, the solution to the integral [tex]\(\int x e^{-2x} \, dx\)[/tex] is:
[tex]\[ \boxed{\frac{(-2x - 1)}{4} e^{-2x} + C} \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
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