Answer :
Sure! Let's go through the solution step by step.
### Part 1: General Indefinite Integral
We need to find the general indefinite integral of the function [tex]\( 8t^3 - 6t^{-2} \)[/tex].
[tex]\[ \int (8t^3 - 6t^{-2}) \, dt \][/tex]
We'll break this down by integrating each term separately.
1. For the first term [tex]\( 8t^3 \)[/tex]:
[tex]\[ \int 8t^3 \, dt \][/tex]
Using the power rule of integration [tex]\( \int t^n \, dt = \frac{t^{n+1}}{n+1} \)[/tex]:
[tex]\[ \int 8t^3 \, dt = 8 \cdot \frac{t^{3+1}}{3+1} = 8 \cdot \frac{t^4}{4} = 2t^4 \][/tex]
2. For the second term [tex]\( -6t^{-2} \)[/tex]:
[tex]\[ \int -6t^{-2} \, dt \][/tex]
Again using the power rule:
[tex]\[ \int -6t^{-2} \, dt = -6 \cdot \frac{t^{-2+1}}{-2+1} = -6 \cdot \frac{t^{-1}}{-1} = 6t^{-1} = \frac{6}{t} \][/tex]
Summing these results, we get:
[tex]\[ \int (8t^3 - 6t^{-2}) \, dt = 2t^4 + \frac{6}{t} + C \][/tex]
So, the general indefinite integral is:
[tex]\[ 2t^4 + \frac{6}{t} + C \][/tex]
### Part 2: Definite Integral
Now, we need to evaluate the definite integral of the function [tex]\( 8t^3 - 6t^{-2} \)[/tex] from [tex]\( t = 2 \)[/tex] to [tex]\( t = 6 \)[/tex]:
[tex]\[ \int_2^6 (8t^3 - 6t^{-2}) \, dt \][/tex]
We use the antiderivative we found previously and evaluate it at the upper and lower bounds, then take the difference.
Using the antiderivative [tex]\( 2t^4 + \frac{6}{t} \)[/tex]:
1. Evaluate at [tex]\( t = 6 \)[/tex]:
[tex]\[ 2(6)^4 + \frac{6}{6} = 2 \cdot 1296 + 1 = 2592 + 1 = 2593 \][/tex]
2. Evaluate at [tex]\( t = 2 \)[/tex]:
[tex]\[ 2(2)^4 + \frac{6}{2} = 2 \cdot 16 + 3 = 32 + 3 = 35 \][/tex]
Now, subtract the value at the lower bound from the value at the upper bound:
[tex]\[ 2593 - 35 = 2558 \][/tex]
So, the definite integral is:
[tex]\[ \int_2^6 (8t^3 - 6t^{-2}) \, dt = 2558 \][/tex]
### Summary
- The general indefinite integral of [tex]\( 8t^3 - 6t^{-2} \)[/tex] is:
[tex]\[ 2t^4 + \frac{6}{t} + C \][/tex]
- The value of the definite integral of [tex]\( 8t^3 - 6t^{-2} \)[/tex] from [tex]\( t = 2 \)[/tex] to [tex]\( t = 6 \)[/tex] is:
[tex]\[ 2558 \][/tex]
### Part 1: General Indefinite Integral
We need to find the general indefinite integral of the function [tex]\( 8t^3 - 6t^{-2} \)[/tex].
[tex]\[ \int (8t^3 - 6t^{-2}) \, dt \][/tex]
We'll break this down by integrating each term separately.
1. For the first term [tex]\( 8t^3 \)[/tex]:
[tex]\[ \int 8t^3 \, dt \][/tex]
Using the power rule of integration [tex]\( \int t^n \, dt = \frac{t^{n+1}}{n+1} \)[/tex]:
[tex]\[ \int 8t^3 \, dt = 8 \cdot \frac{t^{3+1}}{3+1} = 8 \cdot \frac{t^4}{4} = 2t^4 \][/tex]
2. For the second term [tex]\( -6t^{-2} \)[/tex]:
[tex]\[ \int -6t^{-2} \, dt \][/tex]
Again using the power rule:
[tex]\[ \int -6t^{-2} \, dt = -6 \cdot \frac{t^{-2+1}}{-2+1} = -6 \cdot \frac{t^{-1}}{-1} = 6t^{-1} = \frac{6}{t} \][/tex]
Summing these results, we get:
[tex]\[ \int (8t^3 - 6t^{-2}) \, dt = 2t^4 + \frac{6}{t} + C \][/tex]
So, the general indefinite integral is:
[tex]\[ 2t^4 + \frac{6}{t} + C \][/tex]
### Part 2: Definite Integral
Now, we need to evaluate the definite integral of the function [tex]\( 8t^3 - 6t^{-2} \)[/tex] from [tex]\( t = 2 \)[/tex] to [tex]\( t = 6 \)[/tex]:
[tex]\[ \int_2^6 (8t^3 - 6t^{-2}) \, dt \][/tex]
We use the antiderivative we found previously and evaluate it at the upper and lower bounds, then take the difference.
Using the antiderivative [tex]\( 2t^4 + \frac{6}{t} \)[/tex]:
1. Evaluate at [tex]\( t = 6 \)[/tex]:
[tex]\[ 2(6)^4 + \frac{6}{6} = 2 \cdot 1296 + 1 = 2592 + 1 = 2593 \][/tex]
2. Evaluate at [tex]\( t = 2 \)[/tex]:
[tex]\[ 2(2)^4 + \frac{6}{2} = 2 \cdot 16 + 3 = 32 + 3 = 35 \][/tex]
Now, subtract the value at the lower bound from the value at the upper bound:
[tex]\[ 2593 - 35 = 2558 \][/tex]
So, the definite integral is:
[tex]\[ \int_2^6 (8t^3 - 6t^{-2}) \, dt = 2558 \][/tex]
### Summary
- The general indefinite integral of [tex]\( 8t^3 - 6t^{-2} \)[/tex] is:
[tex]\[ 2t^4 + \frac{6}{t} + C \][/tex]
- The value of the definite integral of [tex]\( 8t^3 - 6t^{-2} \)[/tex] from [tex]\( t = 2 \)[/tex] to [tex]\( t = 6 \)[/tex] is:
[tex]\[ 2558 \][/tex]