Evaluate the definite integral. (Round answer to 2 decimal places.)
[tex]\int_2^5 \frac{8 x^2+7}{\sqrt{x}} d x=79.95[/tex]
Answer (2)
To evaluate the definite integral ∫ 2 5 x 8 x 2 + 7 d x , we rewrite the integrand, find the antiderivative, and apply the limits of integration. After computing and rounding, the answer is 172.29. This represents the accumulated area under the curve of the function from 2 to 5.
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Rewrite the integrand: x 8 x 2 + 7 = 8 x 2 3 + 7 x − 2 1 .
Find the antiderivative: ∫ ( 8 x 2 3 + 7 x − 2 1 ) d x = 5 16 x 2 5 + 14 x 2 1 + C .
Evaluate the definite integral: [ 5 16 x 2 5 + 14 x 2 1 ] 2 5 = 94 5 − 26.8 2 .
Approximate and round to two decimal places: 172.29 .
Explanation
Problem Setup We are asked to evaluate the definite integral ∫ 2 5 x 8 x 2 + 7 d x and round the answer to 2 decimal places.
Rewriting the Integrand First, we rewrite the integrand as a sum of power functions: x 8 x 2 + 7 = 8 x 2 − 2 1 + 7 x − 2 1 = 8 x 2 3 + 7 x − 2 1
Finding the Antiderivative Next, we find the antiderivative of the integrand: ∫ ( 8 x 2 3 + 7 x − 2 1 ) d x = 8 ∫ x 2 3 d x + 7 ∫ x − 2 1 d x = 8 ⋅ 2 5 x 2 5 + 7 ⋅ 2 1 x 2 1 + C = 5 16 x 2 5 + 14 x 2 1 + C
Evaluating the Definite Integral Now, we evaluate the definite integral by applying the Fundamental Theorem of Calculus: ∫ 2 5 ( 8 x 2 3 + 7 x − 2 1 ) d x = [ 5 16 x 2 5 + 14 x 2 1 ] 2 5 = ( 5 16 ( 5 ) 2 5 + 14 ( 5 ) 2 1 ) − ( 5 16 ( 2 ) 2 5 + 14 ( 2 ) 2 1 )
Simplifying the Expression We simplify the expression: ( 5 16 ( 5 ) 2 5 + 14 ( 5 ) 2 1 ) − ( 5 16 ( 2 ) 2 5 + 14 ( 2 ) 2 1 ) = ( 5 16 ⋅ 25 5 + 14 5 ) − ( 5 16 ⋅ 4 2 + 14 2 ) = ( 80 5 + 14 5 ) − ( 5 64 2 + 14 2 ) = 94 5 − ( 12.8 2 + 14 2 ) = 94 5 − 26.8 2
Approximating the Result Using a calculator, we approximate the result: 94 5 − 26.8 2 ≈ 94 ( 2.236 ) − 26.8 ( 1.414 ) ≈ 210.2 − 37.9 ≈ 172.3 A more precise calculation gives us 172.2894664133813. Rounding this to two decimal places, we get 172.29.
Final Answer Therefore, the value of the definite integral, rounded to two decimal places, is 172.29.
Examples
Definite integrals are used to calculate areas, volumes, and other accumulated quantities. For example, if you have a function that represents the rate of water flow into a tank, the definite integral of that function over a specific time interval will give you the total amount of water that entered the tank during that time. This is useful in many engineering and physics applications.
