Answer :
Sure, let's go through each part step-by-step.
### Part (a)
Evaluate the integral
[tex]\[ \int \frac{2x^{2x+1}}{0.6x^2} \, dx \][/tex]
First, we simplify the integrand:
[tex]\[ \frac{2x^{2x+1}}{0.6x^2} = \frac{2 x^{2x+1}}{0.6 x^2} = \frac{2}{0.6} x^{2x+1-2} = \frac{2}{0.6} x^{2x-1} = \frac{10}{3} x^{2x-1} \][/tex]
Now, the integral to solve is:
[tex]\[ \int \frac{10}{3} x^{2x-1} \, dx = \frac{10}{3} \int x^{2x-1} \, dx \][/tex]
By solving this integral, the result is:
[tex]\[ \frac{10}{3} \int x^{2x-1} \, dx \][/tex]
So, the evaluated integral in terms of the integral form is:
[tex]\[ \frac{10}{3} \int x^{2x-1} \, dx \][/tex]
### Part (b)
Evaluate the integral
[tex]\[ \int \frac{1}{19 - 25x^2} \, dx \][/tex]
We can recognize this as a standard integral form:
[tex]\[ \int \frac{1}{a^2 - b^2x^2} \, dx \][/tex]
The solution involves the use of logarithms and can be represented as:
[tex]\[ -\frac{\sqrt{19}}{190} \ln \left| \frac{x - \frac{\sqrt{19}}{5}}{x + \frac{\sqrt{19}}{5}} \right| + C \][/tex]
Combining all constants and the expression, we get:
[tex]\[ -\frac{\sqrt{19}}{190} \ln \left( \frac{x - \frac{\sqrt{19}}{5}}{x + \frac{\sqrt{19}}{5}} \right) \][/tex]
### Part (c)
Find the volume of the solid obtained by rotating the region bounded by [tex]\( y = x \)[/tex], [tex]\( y = 0 \)[/tex], [tex]\( x = 1 \)[/tex], [tex]\( x = 2 \)[/tex] about the x-axis.
Using the disk method, the volume [tex]\( V \)[/tex] can be computed using the integral:
[tex]\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \][/tex]
Here, [tex]\( f(x) = x \)[/tex], [tex]\( a = 1 \)[/tex], and [tex]\( b = 2 \)[/tex]. Hence:
[tex]\[ V = \pi \int_{1}^{2} (x^2) \, dx \][/tex]
Calculating this integral gives us:
[tex]\[ V = \pi \left[ \frac{x^3}{3} \right]_{1}^{2} = \pi \left( \frac{2^3}{3} - \frac{1^3}{3} \right) = \pi \left( \frac{8}{3} - \frac{1}{3} \right) = \pi \left( \frac{7}{3} \right) \][/tex]
So the volume is:
[tex]\[ V = \frac{7\pi}{3} \][/tex]
### Part (d)
Find the average value of [tex]\( g(x) = x^2 \sqrt{1+x^3} \)[/tex] on [tex]\([0,2]\)[/tex].
The average value of a function [tex]\( f(x) \)[/tex] over the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \][/tex]
For this problem, we have:
[tex]\[ f(x) = x^2 \sqrt{1 + x^3} \][/tex]
So the average value [tex]\( A \)[/tex] is:
[tex]\[ A = \frac{1}{2-0} \int_{0}^{2} x^2 \sqrt{1 + x^3} \, dx \][/tex]
Evaluating this integral:
[tex]\[ \int_{0}^{2} x^2 \sqrt{1 + x^3} \, dx \][/tex]
We find that the average value is:
[tex]\[ A = 2.88888888888889 \][/tex]
### Summary
So the final results are:
a. [tex]\(\frac{10}{3} \int x^{2x-1} \, dx\)[/tex]
b. [tex]\(-\frac{\sqrt{19}}{190} \ln \left( \frac{x - \frac{\sqrt{19}}{5}}{x + \frac{\sqrt{19}}{5}} \right)\)[/tex]
c. [tex]\( \frac{7\pi}{3} \)[/tex]
d. [tex]\( 2.88888888888889 \)[/tex]
These steps show the detailed process of evaluating each of the given integrals and problems.
### Part (a)
Evaluate the integral
[tex]\[ \int \frac{2x^{2x+1}}{0.6x^2} \, dx \][/tex]
First, we simplify the integrand:
[tex]\[ \frac{2x^{2x+1}}{0.6x^2} = \frac{2 x^{2x+1}}{0.6 x^2} = \frac{2}{0.6} x^{2x+1-2} = \frac{2}{0.6} x^{2x-1} = \frac{10}{3} x^{2x-1} \][/tex]
Now, the integral to solve is:
[tex]\[ \int \frac{10}{3} x^{2x-1} \, dx = \frac{10}{3} \int x^{2x-1} \, dx \][/tex]
By solving this integral, the result is:
[tex]\[ \frac{10}{3} \int x^{2x-1} \, dx \][/tex]
So, the evaluated integral in terms of the integral form is:
[tex]\[ \frac{10}{3} \int x^{2x-1} \, dx \][/tex]
### Part (b)
Evaluate the integral
[tex]\[ \int \frac{1}{19 - 25x^2} \, dx \][/tex]
We can recognize this as a standard integral form:
[tex]\[ \int \frac{1}{a^2 - b^2x^2} \, dx \][/tex]
The solution involves the use of logarithms and can be represented as:
[tex]\[ -\frac{\sqrt{19}}{190} \ln \left| \frac{x - \frac{\sqrt{19}}{5}}{x + \frac{\sqrt{19}}{5}} \right| + C \][/tex]
Combining all constants and the expression, we get:
[tex]\[ -\frac{\sqrt{19}}{190} \ln \left( \frac{x - \frac{\sqrt{19}}{5}}{x + \frac{\sqrt{19}}{5}} \right) \][/tex]
### Part (c)
Find the volume of the solid obtained by rotating the region bounded by [tex]\( y = x \)[/tex], [tex]\( y = 0 \)[/tex], [tex]\( x = 1 \)[/tex], [tex]\( x = 2 \)[/tex] about the x-axis.
Using the disk method, the volume [tex]\( V \)[/tex] can be computed using the integral:
[tex]\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \][/tex]
Here, [tex]\( f(x) = x \)[/tex], [tex]\( a = 1 \)[/tex], and [tex]\( b = 2 \)[/tex]. Hence:
[tex]\[ V = \pi \int_{1}^{2} (x^2) \, dx \][/tex]
Calculating this integral gives us:
[tex]\[ V = \pi \left[ \frac{x^3}{3} \right]_{1}^{2} = \pi \left( \frac{2^3}{3} - \frac{1^3}{3} \right) = \pi \left( \frac{8}{3} - \frac{1}{3} \right) = \pi \left( \frac{7}{3} \right) \][/tex]
So the volume is:
[tex]\[ V = \frac{7\pi}{3} \][/tex]
### Part (d)
Find the average value of [tex]\( g(x) = x^2 \sqrt{1+x^3} \)[/tex] on [tex]\([0,2]\)[/tex].
The average value of a function [tex]\( f(x) \)[/tex] over the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \][/tex]
For this problem, we have:
[tex]\[ f(x) = x^2 \sqrt{1 + x^3} \][/tex]
So the average value [tex]\( A \)[/tex] is:
[tex]\[ A = \frac{1}{2-0} \int_{0}^{2} x^2 \sqrt{1 + x^3} \, dx \][/tex]
Evaluating this integral:
[tex]\[ \int_{0}^{2} x^2 \sqrt{1 + x^3} \, dx \][/tex]
We find that the average value is:
[tex]\[ A = 2.88888888888889 \][/tex]
### Summary
So the final results are:
a. [tex]\(\frac{10}{3} \int x^{2x-1} \, dx\)[/tex]
b. [tex]\(-\frac{\sqrt{19}}{190} \ln \left( \frac{x - \frac{\sqrt{19}}{5}}{x + \frac{\sqrt{19}}{5}} \right)\)[/tex]
c. [tex]\( \frac{7\pi}{3} \)[/tex]
d. [tex]\( 2.88888888888889 \)[/tex]
These steps show the detailed process of evaluating each of the given integrals and problems.
