Logo

AskSia

Plus

result Spring 2012 11. The integral ∫1∞x2e−2xdx \int_{1}^{\infty} x^{2} e^{-2 x} d x i...
Jul 17, 2024
result
Solution by Steps
step 1
We need to evaluate the definite integral from 1 to infinity of x2e−2x x^2 e^{-2x} with respect to x x
step 2
The integral is given by: ∫1∞x2e−2x dx \int_{1}^{\infty} x^2 e^{-2x} \, dx We can use integration by parts to solve this. Let u=x2 u = x^2 and dv=e−2x dx dv = e^{-2x} \, dx
step 3
Compute du du and v v : du=2x dx du = 2x \, dx v=∫e−2x dx=−12e−2x v = \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x}
step 4
Apply integration by parts formula: ∫u dv=uv−∫v du \int u \, dv = uv - \int v \, du ∫x2e−2x dx=x2(−12e−2x)−∫(−12e−2x)(2x) dx \int x^2 e^{-2x} \, dx = x^2 \left( -\frac{1}{2} e^{-2x} \right) - \int \left( -\frac{1}{2} e^{-2x} \right) (2x) \, dx =−12x2e−2x+∫xe−2x dx = -\frac{1}{2} x^2 e^{-2x} + \int x e^{-2x} \, dx
step 5
Now, we need to integrate xe−2x x e^{-2x} . Let u=x u = x and dv=e−2x dx dv = e^{-2x} \, dx : du=dx du = dx v=−12e−2x v = -\frac{1}{2} e^{-2x} ∫xe−2x dx=x(−12e−2x)−∫(−12e−2x) dx \int x e^{-2x} \, dx = x \left( -\frac{1}{2} e^{-2x} \right) - \int \left( -\frac{1}{2} e^{-2x} \right) \, dx =−12xe−2x+14e−2x = -\frac{1}{2} x e^{-2x} + \frac{1}{4} e^{-2x}
step 6
Substitute back into the original integral: ∫x2e−2x dx=−12x2e−2x+(−12xe−2x+14e−2x) \int x^2 e^{-2x} \, dx = -\frac{1}{2} x^2 e^{-2x} + \left( -\frac{1}{2} x e^{-2x} + \frac{1}{4} e^{-2x} \right) =−12e−2x(x2+x+12) = -\frac{1}{2} e^{-2x} (x^2 + x + \frac{1}{2})
step 7
Evaluate the definite integral from 1 to infinity: [−12e−2x(x2+x+12)]1∞ \left[ -\frac{1}{2} e^{-2x} (x^2 + x + \frac{1}{2}) \right]_{1}^{\infty} As x→∞ x \to \infty , e−2x→0 e^{-2x} \to 0 , so the term at infinity is 0. Evaluate at x=1 x = 1 : −12e−2(12+1+12)=−12e−2(52)=−54e−2 -\frac{1}{2} e^{-2} (1^2 + 1 + \frac{1}{2}) = -\frac{1}{2} e^{-2} \left( \frac{5}{2} \right) = -\frac{5}{4} e^{-2}
Answer
54e−2 \frac{5}{4} e^{-2}
Key Concept
Definite Integral Evaluation
Explanation
The integral of x2e−2x x^2 e^{-2x} from 1 to infinity is evaluated using integration by parts twice, resulting in 54e−2 \frac{5}{4} e^{-2} .
}
Generate me a similar question
Explain how to calculate the definite integral ∫0∞x3e−3x \int_{0}^{\infty} x^3 e^{-3x}
© 2023 AskSia.AI all rights reserved