Answer :
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]
