Let’s break down and solve each part of this exercise step by step.

1. (a) Show that $\int_1^{+\infty} e^{-x} \, dx$ is convergent and calculate its value.

Solution:

The integral is: $\int_1^{+\infty} e^{-x} \, dx.$

The general formula for the integral of $e^{-x}$ is: $\int e^{-x} \, dx = -e^{-x} + C.$

Evaluate the definite integral: $\int_1^{+\infty} e^{-x} \, dx = \left[-e^{-x}\right]_1^{+\infty}.$

1. As $x \to +\infty$, $e^{-x} \to 0$.
2. At $x = 1$, $-e^{-x} = -e^{-1}$.

Thus: $\int_1^{+\infty} e^{-x} \, dx = 0 - (-e^{-1}) = e^{-1}.$

The value is: $\int_1^{+\infty} e^{-x} \, dx = \frac{1}{e}.$

1. (b) Deduce that $\int_1^{+\infty} e^{-x^2} \, dx$ is convergent.

Solution:

The function $e^{-x^2}$ decreases faster than $e^{-x}$ as $x \to +\infty$. Since $\int_1^{+\infty} e^{-x} \, dx$ converges, by comparison, $\int_1^{+\infty} e^{-x^2} \, dx$ also converges by the comparison test.

2. (a) Show that $\int_0^1 \ln x \, dx$ is convergent and calculate its value.

Solution:

The integral is: $I = \int_0^1 \ln x \, dx.$

Using integration by parts: $\int \ln x \, dx = x \ln x - \int x \, \frac{1}{x} \, dx = x \ln x - x + C.$

Evaluate the definite integral: $I = \left[x \ln x - x\right]_0^1.$

1. At $x = 1$: $x \ln x - x = 1 \ln 1 - 1 = -1.$

2. At $x = 0$: $\lim_{x \to 0^+} (x \ln x) = \lim_{x \to 0^+} \frac{\ln x}{1/x} = 0 \quad \text{(using L'Hopital's Rule)}.$

Thus: $I = -1 - 0 = -1.$

The value is: $I = -1.$

2. (b) Show that $\int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx$ is convergent and calculate its value.

Solution:

Substitute $x = \sin t$, so $dx = \cos t \, dt$, and $\sqrt{1-x^2} = \cos t$. The limits transform as:

• When $x = 0$, $t = 0$.
• When $x = 1$, $t = \frac{\pi}{2}$.

The integral becomes: $J = \int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx = \int_0^{\pi/2} \frac{\sin^2 t}{\cos t} \cos t \, dt.$

Simplify: $J = \int_0^{\pi/2} \sin^2 t \, dt.$

Use the identity $\sin^2 t = \frac{1 - \cos(2t)}{2}$: $J = \int_0^{\pi/2} \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \int_0^{\pi/2} 1 \, dt - \frac{1}{2} \int_0^{\pi/2} \cos(2t) \, dt.$

1. Compute $\int_0^{\pi/2} 1 \, dt$: $\int_0^{\pi/2} 1 \, dt = \frac{\pi}{2}.$

2. Compute $\int_0^{\pi/2} \cos(2t) \, dt$: $\int \cos(2t) \, dt = \frac{\sin(2t)}{2}.$ Evaluate from $0$ to $\frac{\pi}{2}$: $\int_0^{\pi/2} \cos(2t) \, dt = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0.$

Thus: $J = \frac{1}{2} \left(\frac{\pi}{2} - 0\right) = \frac{\pi}{4}.$

The value is: $J = \frac{\pi}{4}.$

Summary of Results:

1. (a) $\int_1^{+\infty} e^{-x} \, dx = \frac{1}{e}$.
2. (b) $\int_1^{+\infty} e^{-x^2} \, dx$ is convergent.
3. (a) $\int_0^1 \ln x \, dx = -1$.
4. (b) $\int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx = \frac{\pi}{4}$.

Would you like further details on any step or explanation? Here are some related questions to consider:

1. How does the comparison test for convergence work?
2. What are the conditions for applying integration by parts?
3. Why does the substitution $x = \sin t$ simplify integrals involving $\sqrt{1-x^2}$?
4. Can $e^{-x^2}$ be integrated analytically over $[1, +\infty)$?
5. How does L'Hopital's Rule apply to limits like $\lim_{x \to 0^+} x \ln x$?

Tip: Always check for the behavior of the integrand at the boundaries to assess convergence!