6I.X Excercises - JulTob/Mathematics GitHub Wiki

⭐️ Let’s calculate an integral!

I = \int \int_D \frac{1}{\sqrt[4]{x^2 \cdot y}} \, dx \, dy,

where $( D = { (x, y) : x \in [0, 1], y \in [0, 1] } )$.

We will proceed step by step, carefully explaining the theory involved.

Step 1: Understanding the integral and region $( D )$

The integral is defined over the region $( D )$, which is a square in the $( xy )$-plane with $( x \in [0, 1] )$)$ and $( y \in [0, 1] )$.

xychart-beta
    title "D region"

    x-axis "x" [0, 1, 1, 0]
    y-axis "y" 0 --> 1 
    line[0, 0, 1, 1]

The function to be integrated is:

f(x, y) = \frac{1}{\sqrt[4]{x^2 \cdot y}} = \color{lime}{(x^2 \cdot y)^{-1/4}}

The double integral can be expressed explicitly as:

I 
= \color{gold} \displaystyle \lim_{k \to 0} \displaystyle \lim_{l \to 0} {\int_k^1 \int_l^1} \color{silver} \frac{1}{\sqrt[4]{x^2 \cdot y}} \, dx \, dy.
= \color{gold}{\int_0^1 \int_0^1} \color{silver} \frac{1}{\sqrt[4]{x^2 \cdot y}} \, dx \, dy.

Step 2: Separating the function $( f(x, y) )$

Notice that $( f(x, y) = \frac{1}{\sqrt[4]{x^2 \cdot y}} )$ can be written as a product of functions of $( x )$ and $( y )$:

 f(x, y) = (x^2 \cdot y)^{\color{lime}-1/4} = \color{lime} x^{-1/2} \cdot y^{-1/4}.

This separation is valid because multiplication is commutative and $( x^2 \cdot y )$ is a product of two independent variables.

Step 3: Fubini’s theorem

Because the region $( D )$ is a rectangle, the limits of integration for $( x )$ and $( y )$ are independent. Therefore, we can use Fubini's theorem, which allows us to write:

I = \int_0^1 \int_0^1 f(x, y) \, dx \, dy = \int_0^1 \left( \int_0^1 x^{-1/2} \cdot y^{-1/4} \, dx \right) dy.

Since $( y^{-1/4} )$ is independent of $( x )$, it can be factored out of the inner integral:

I = { \color{cyan} \int_0^1 y^{-1/4}  \left( {\color{orange} \int_0^1 x^{-1/2} \, dx} \right) dy }.

Step 4: Evaluating the inner integral over $( x )$

We now calculate the inner integral:

\color{orange}
\int_0^1 x^{-1/2} \, dx.

The function $( x^{-1/2} )$ is integrable on $( [0, 1] )$. The integral is computed as:

\color{orange}
\int_0^1 x^{-1/2} \, dx = \int_0^1 x^{-\frac{1}{2}} \, dx.

The antiderivative of $( x^{-1/2} )$ is:

\color{orange}
\int x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}} + C,

where $( C )$ is the constant of integration. Evaluating this from $( 0 )$ to $( 1 )$:

\color{orange}
\int_0^1 x^{-1/2} \, dx = 2 \cdot \left[ x^{1/2} \right]_0^1 = 2 \cdot (1 - 0) = 2.

Thus, the inner integral simplifies to:

\color{orange}
\int_0^1 x^{-1/2} \, dx = 2.

Step 5: Substituting into the outer integral

Substitute $( \int_0^1 x^{-1/2} , dx = 2 )$ back into the expression for $( I )$:

\color{cyan}
I = \int_0^1 y^{-1/4} \cdot {\color{orange}2} \, dy.

Factor out the constant $( 2 )$:

\color{cyan}
I = 2 \int_0^1 y^{-1/4} \, dy.

Step 6: Evaluating the outer integral over $( y )$

We now calculate the integral:

\color{cyan}
\int_0^1 y^{-1/4} \, dy.

The function $( y^{-1/4} )$ is integrable on $( [0, 1] )$. The antiderivative of $( y^{-1/4} )$ is:

\color{cyan}
\int y^{-1/4} \, dy = \frac{y^{3/4}}{3/4} {\color{crimson}+ C} = \frac{4}{3} y^{3/4} {\color{crimson}+ C}.

Evaluating this from $( 0 )$ to $( 1 )$:

\color{cyan}
\int_0^1 y^{-1/4} \, dy = \frac{4}{3} \cdot \left[ y^{3/4} \right]_0^1 = \frac{4}{3} \cdot (1 - 0) = \frac{4}{3}.

Step 7: Final result

Substitute $( \int_0^1 y^{-1/4} , dy = \frac{4}{3} )$ back into the expression for $( I )$:

\color{lime}
I = 2 \cdot \frac{4}{3} = \frac{8}{3}.

Thus, the value of the double integral is:

\color{lime}
\boxed{\boxed{\frac{8}{3}}}

⭐️ Let’s calculate an integral!

xychart-beta
    title "D region"

    x-axis "x" [0, 1, 1]
    y-axis "y" 0 --> 1 
    line[0,  1, 0]

I = \int \int_D \frac{1}{\sqrt{|x \cdot y|}} \, dx \, dy,

where $( D = {(x, y) : x \in [0, 1], y \in [0, 1], y \leq x} )$.

Step 1: Understanding the region $( D )$

The region $( D )$ is defined by:

  1. $( x \in [0, 1] )$,
  2. $( y \in [0, 1] )$,
  3. $( y \leq x )$.

This means $( D )$ is the triangular region in the first quadrant where $( y \leq x )$ and both $( x )$ and $( y )$ are bounded by $( [0, 1] )$. In terms of limits:

  • For a fixed $( x )$, $( y )$ varies from $( 0 )$ to $( x )$.
  • $( x )$ varies from $( 0 )$ to $( 1 )$.

The integral becomes:

I = \int_0^1 \int_0^x \frac{1}{\sqrt{x \cdot y}} \, dy \, dx.

Step 2: Separating the square root

The integrand $( \frac{1}{\sqrt{x \cdot y}} )$ can be rewritten as:

\frac{1}{\sqrt{x \cdot y}} = \frac{1}{\sqrt{x}} \cdot \frac{1}{\sqrt{y}}.

Thus, the integral becomes:

I = \int_0^1 \frac{1}{\sqrt{x}} \left( \int_0^x \frac{1}{\sqrt{y}} \, dy \right) dx.

Step 3: Evaluating the inner integral over $( y )$

We calculate the inner integral:

\int_0^x \frac{1}{\sqrt{y}} \, dy.

The antiderivative of $( y^{-1/2} )$ is:

\int y^{-1/2} \, dy = 2y^{1/2} + C.

Evaluating from $( 0 )$ to $( x )$:

\int_0^x \frac{1}{\sqrt{y}} \, dy = 2\sqrt{x} - 2\sqrt{0} = 2\sqrt{x}.

Substitute this back into the integral:

I = \int_0^1 \frac{1}{\sqrt{x}} \cdot 2\sqrt{x} \, dx.

Step 4: Simplifying the integrand

Simplify $( \frac{1}{\sqrt{x}} \cdot 2\sqrt{x} )$:

\frac{1}{\sqrt{x}} \cdot 2\sqrt{x} = 2.

Thus, the integral becomes:

I = \int_0^1 2 \, dx.

Step 5: Evaluating the final integral

The integral of $( 2 )$ with respect to $( x )$)$ is:

\int_0^1 2 \, dx = 2x \Big|_0^1 = 2(1) - 2(0) = 2.

Final Answer

The value of the integral is:

\boxed{2}.

⭐️ Let's solve the double integral:

\int_0^1 \int_0^{e^y} \log(x) \, dx \, dy.

    xychart-beta
    title "Integration Region"
    x-axis "x" 0 --> 3 
    x-axis [ 0.8631578947368421, 1.0157894736842106, 1.168421052631579, 1.3210526315789475, 1.4736842105263157, 1.6263157894736844, 1.7789473684210526, 1.931578947368421, 2.0842105263157893, 2.236842105263158, 2.389473684210526, 2.542105263157895, 2.694736842105263, 2.8473684210526313, 3.0]
    y-axis "log(x)" 0 --> 1.10
    line [ -0.1471576443362877, 0.015666116744399456, 0.15565330971179336, 0.2784288669712978, 0.38776553100876343, 0.4863172047372594, 0.5760218233225326, 0.6583377758940842, 0.7343901390939943, 0.8050650967639307, 0.8710731258808618, 0.9329925814920866, 0.9913005528790216, 1.0463952066859954, 1.0986122886681098]


    line "Boundary x=0" [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
    line "Boundary x=1" [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

Step 1: Analyzing the region of integration

The integration bounds are:

  • $( y \in [0, 1] )$,
  • $( x \in [0, e^y] )$.

For a fixed $( y )$, $( x )$ ranges from $( 0 )$ to $( e^y )$. This means we first integrate with respect to $( x )$, keeping $( y )$ constant, and then integrate with respect to $( y )$.

Step 2: Setting up the integral

The integral is expressed as:

\int_0^1 \int_0^{e^y} \log(x) \, dx \, dy.

Step 3: Integrating with respect to $( x )$

The inner integral is:

\int_0^{e^y} \log(x) \, dx.

To integrate $( \log(x) )$, we use integration by parts. Recall the formula:

\int \log(x) \, dx = x \log(x) - x + C.

Applying this to the definite integral:

\int_0^{e^y} \log(x) \, dx = \left[ x \log(x) - x \right]_0^{e^y}.

For $( x = e^y )$:

x \log(x) - x = e^y \log(e^y) - e^y = e^y \cdot y - e^y.

For $( x = 0 )$, $( \log(x) )$ is undefined, but we take the limit as $( x \to 0^+ )$. Since $( x \log(x) \to 0 )$ as $( x \to 0^+ )$, the lower bound contributes $( 0 )$.

Thus, the inner integral evaluates to:

\int_0^{e^y} \log(x) \, dx = e^y y - e^y.

Step 4: Substituting into the outer integral

Substitute the result into the outer integral:

\int_0^1 \int_0^{e^y} \log(x) \, dx \, dy = \int_0^1 \left( e^y y - e^y \right) \, dy.

Simplify:

\int_0^1 \left( e^y y - e^y \right) \, dy = \int_0^1 e^y y \, dy - \int_0^1 e^y \, dy.

Step 5: Evaluating $( \int_0^1 e^y , dy )$

The integral of $( e^y )$ is straightforward:

\int e^y \, dy = e^y + C.

Evaluate from $( 0 )$ to $( 1 )$:

\int_0^1 e^y \, dy = \left[ e^y \right]_0^1 = e^1 - e^0 = e - 1.

Step 6: Evaluating $( \int_0^1 e^y y , dy )$

For $( \int_0^1 e^y y , dy )$, we use integration by parts. Recall the formula:

\int u \, dv = uv - \int v \, du.

Let:

  • $( u = y )$ (so $( du = dy )$),
  • $( dv = e^y , dy )$ (so $( v = e^y )$).

Then:

\int e^y y \, dy = y e^y - \int e^y \, dy.

Substitute $( \int e^y , dy = e^y )$:

\int e^y y \, dy = y e^y - e^y + C.

Evaluate from $( 0 )$ to $( 1 )$:

\int_0^1 e^y y \, dy = \left[ y e^y - e^y \right]_0^1.

For $( y = 1 )$:

y e^y - e^y = 1 \cdot e - e = 0.

For $( y = 0 )$:

y e^y - e^y = 0 \cdot e - e = 0.

The value of $( \int_0^1 =

Let me carefully re-check and finalize the evaluation of the integral. We aim to compute:

\int_0^1 e^y y \, dy - \int_0^1 e^y \, dy.

Step 6 (continued): Correct evaluation of $( \int_0^1 e^y y , dy )$

From integration by parts, the result was:

\int e^y y \, dy = y e^y - \int e^y \, dy.

Substitute $( \int e^y , dy = e^y )$:

\int e^y y \, dy = y e^y - e^y.

Now evaluate from $( 0 )$ to $( 1 )$:

  1. At $( y = 1 )$:
   y e^y - e^y = 1 \cdot e - e = 0.
  1. At $( y = 0 )$:
   y e^y - e^y = 0 \cdot e - e = -e.

So:

\int_0^1 e^y y \, dy = \left[ y e^y - e^y \right]_0^1 = \big(0 - (-e)\big) = e.

Step 7: Substituting all results back

Now substitute the two results:

  1. $( \int_0^1 e^y y , dy = e )$,
  2. $( \int_0^1 e^y , dy = e - 1 )$.

Thus:

\int_0^1 \left( e^y y - e^y \right) \, dy = e - (e - 1).

Simplify:

e - (e - 1) = 1.

Final Answer

The value of the integral is:

\boxed{1}.