Evaluate the definite integral [tex]$\int_2^6(8 x+5) d x$[/tex]
Answer (2)
Find the antiderivative of the function: 4 x 2 + 5 x .
Evaluate the antiderivative at the upper limit (6): 4 ( 6 ) 2 + 5 ( 6 ) = 174 .
Evaluate the antiderivative at the lower limit (2): 4 ( 2 ) 2 + 5 ( 2 ) = 26 .
Subtract the value at the lower limit from the value at the upper limit: 174 − 26 = 148 .
Explanation
Understanding the Problem We are asked to evaluate the definite integral ∫ 2 6 ( 8 x + 5 ) d x . This means we need to find the antiderivative of the function 8 x + 5 , and then evaluate it at the limits of integration, 6 and 2, and subtract the value at 2 from the value at 6.
Finding the Antiderivative First, let's find the antiderivative of 8 x + 5 . Using the power rule for integration, the antiderivative of 8 x is 4 x 2 , and the antiderivative of 5 is 5 x . So the antiderivative of 8 x + 5 is 4 x 2 + 5 x + C , where C is the constant of integration. Since we are evaluating a definite integral, the constant of integration will cancel out, so we don't need to worry about it.
Evaluating at the Upper Limit Now, we need to evaluate the antiderivative at the upper limit of integration, which is 6. So we have 4 ( 6 ) 2 + 5 ( 6 ) = 4 ( 36 ) + 30 = 144 + 30 = 174 .
Evaluating at the Lower Limit Next, we need to evaluate the antiderivative at the lower limit of integration, which is 2. So we have 4 ( 2 ) 2 + 5 ( 2 ) = 4 ( 4 ) + 10 = 16 + 10 = 26 .
Calculating the Definite Integral Finally, we subtract the value of the antiderivative at the lower limit from the value at the upper limit: 174 − 26 = 148 . Therefore, the definite integral ∫ 2 6 ( 8 x + 5 ) d x = 148 .
Final Answer The definite integral ∫ 2 6 ( 8 x + 5 ) d x evaluates to 148.
Examples
Definite integrals are used to calculate the area under a curve. For example, if you have a function that represents the velocity of an object over time, the definite integral of that function over a certain time interval will give you the displacement of the object during that time interval. This is useful in physics and engineering for calculating distances, work, and other quantities.
The definite integral ∫ 2 6 ( 8 x + 5 ) d x is evaluated by finding the antiderivative and then calculating the difference between the values at the upper and lower limits. The result is 148. This represents the area under the curve of the function from x = 2 to x = 6.
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