Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Certainly! Let's look at these improper integrals step-by-step.
### Part (a) [tex]\(\int_0^{\infty} \frac{e^{-x}}{1+e^{-x}} \,dx\)[/tex]
1. We begin with the integrand function:
[tex]\[ \frac{e^{-x}}{1 + e^{-x}} \][/tex]
2. This integrand can be simplified by noting a substitution. Let [tex]\( u = e^{-x} \)[/tex]. Hence, [tex]\( du = -e^{-x} \, dx = -u \, dx \)[/tex]. When [tex]\( x \to 0 \)[/tex], [tex]\( u \to 1 \)[/tex], and when [tex]\( x \to \infty \)[/tex], [tex]\( u \to 0 \)[/tex].
3. Rewrite the integral in terms of [tex]\( u \)[/tex]:
[tex]\[ \int_0^{\infty} \frac{e^{-x}}{1 + e^{-x}} \, dx = \int_1^{0} \frac{u}{1+u} \cdot \left( -\frac{1}{u} \, du \right) = \int_1^0 \frac{1}{1+u} \, du \][/tex]
4. Switching the limits of integration changes the sign:
[tex]\[ \int_1^0 \frac{1}{1+u} \, du = -\int_0^1 \frac{1}{1+u} \, du \][/tex]
5. This becomes a standard integral of the form:
[tex]\[ \int_0^1 \frac{1}{1+u} \, du = [ \log(1+u) ]_0^1 \][/tex]
6. Evaluating this definite integral:
[tex]\[ \log(1+1) - \log(1+0) = \log(2) - \log(1) = \log(2) \][/tex]
Thus, the result for part (a) is:
[tex]\[ \int_0^{\infty} \frac{e^{-x}}{1+e^{-x}} \,dx = \log(2) \][/tex]
### Part (b) [tex]\(\int_1^2 \frac{1}{\sqrt{x-1}} \,dx\)[/tex]
1. We start with the integrand function:
[tex]\[ \frac{1}{\sqrt{x-1}} \][/tex]
2. This is an improper integral because the integrand becomes infinite at the lower limit of integration [tex]\( x = 1 \)[/tex]. We can solve it by making a substitution [tex]\( u = x - 1 \)[/tex]. Hence, [tex]\( du = dx \)[/tex]. When [tex]\( x \to 1 \)[/tex], [tex]\( u \to 0 \)[/tex], and when [tex]\( x \to 2 \)[/tex], [tex]\( u \to 1 \)[/tex].
3. Rewrite the integral in terms of [tex]\( u \)[/tex]:
[tex]\[ \int_1^2 \frac{1}{\sqrt{x-1}} \, dx = \int_0^1 \frac{1}{\sqrt{u}} \, du \][/tex]
4. This becomes a standard integral:
[tex]\[ \int_0^1 u^{-\frac{1}{2}} \, du \][/tex]
5. Integrate using the power rule:
[tex]\[ \int_0^1 u^{-\frac{1}{2}} \, du = \left[ \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_0^1 = \left[ 2u^{\frac{1}{2}} \right]_0^1 \][/tex]
6. Evaluating this definite integral gives:
[tex]\[ 2(1^{\frac{1}{2}}) - 2(0^{\frac{1}{2}}) = 2(1) - 2(0) = 2 \][/tex]
Thus, the result for part (b) is:
[tex]\[ \int_1^2 \frac{1}{\sqrt{x-1}} \,dx = 2 \][/tex]
### Summary
The values of the improper integrals are:
[tex]\[ \text{(a)} \int_0^{\infty} \frac{e^{-x}}{1+e^{-x}} \,dx = \log(2) \][/tex]
[tex]\[ \text{(b)} \int_1^2 \frac{1}{\sqrt{x-1}} \,dx = 2 \][/tex]
### Part (a) [tex]\(\int_0^{\infty} \frac{e^{-x}}{1+e^{-x}} \,dx\)[/tex]
1. We begin with the integrand function:
[tex]\[ \frac{e^{-x}}{1 + e^{-x}} \][/tex]
2. This integrand can be simplified by noting a substitution. Let [tex]\( u = e^{-x} \)[/tex]. Hence, [tex]\( du = -e^{-x} \, dx = -u \, dx \)[/tex]. When [tex]\( x \to 0 \)[/tex], [tex]\( u \to 1 \)[/tex], and when [tex]\( x \to \infty \)[/tex], [tex]\( u \to 0 \)[/tex].
3. Rewrite the integral in terms of [tex]\( u \)[/tex]:
[tex]\[ \int_0^{\infty} \frac{e^{-x}}{1 + e^{-x}} \, dx = \int_1^{0} \frac{u}{1+u} \cdot \left( -\frac{1}{u} \, du \right) = \int_1^0 \frac{1}{1+u} \, du \][/tex]
4. Switching the limits of integration changes the sign:
[tex]\[ \int_1^0 \frac{1}{1+u} \, du = -\int_0^1 \frac{1}{1+u} \, du \][/tex]
5. This becomes a standard integral of the form:
[tex]\[ \int_0^1 \frac{1}{1+u} \, du = [ \log(1+u) ]_0^1 \][/tex]
6. Evaluating this definite integral:
[tex]\[ \log(1+1) - \log(1+0) = \log(2) - \log(1) = \log(2) \][/tex]
Thus, the result for part (a) is:
[tex]\[ \int_0^{\infty} \frac{e^{-x}}{1+e^{-x}} \,dx = \log(2) \][/tex]
### Part (b) [tex]\(\int_1^2 \frac{1}{\sqrt{x-1}} \,dx\)[/tex]
1. We start with the integrand function:
[tex]\[ \frac{1}{\sqrt{x-1}} \][/tex]
2. This is an improper integral because the integrand becomes infinite at the lower limit of integration [tex]\( x = 1 \)[/tex]. We can solve it by making a substitution [tex]\( u = x - 1 \)[/tex]. Hence, [tex]\( du = dx \)[/tex]. When [tex]\( x \to 1 \)[/tex], [tex]\( u \to 0 \)[/tex], and when [tex]\( x \to 2 \)[/tex], [tex]\( u \to 1 \)[/tex].
3. Rewrite the integral in terms of [tex]\( u \)[/tex]:
[tex]\[ \int_1^2 \frac{1}{\sqrt{x-1}} \, dx = \int_0^1 \frac{1}{\sqrt{u}} \, du \][/tex]
4. This becomes a standard integral:
[tex]\[ \int_0^1 u^{-\frac{1}{2}} \, du \][/tex]
5. Integrate using the power rule:
[tex]\[ \int_0^1 u^{-\frac{1}{2}} \, du = \left[ \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_0^1 = \left[ 2u^{\frac{1}{2}} \right]_0^1 \][/tex]
6. Evaluating this definite integral gives:
[tex]\[ 2(1^{\frac{1}{2}}) - 2(0^{\frac{1}{2}}) = 2(1) - 2(0) = 2 \][/tex]
Thus, the result for part (b) is:
[tex]\[ \int_1^2 \frac{1}{\sqrt{x-1}} \,dx = 2 \][/tex]
### Summary
The values of the improper integrals are:
[tex]\[ \text{(a)} \int_0^{\infty} \frac{e^{-x}}{1+e^{-x}} \,dx = \log(2) \][/tex]
[tex]\[ \text{(b)} \int_1^2 \frac{1}{\sqrt{x-1}} \,dx = 2 \][/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
