Evaluate the integral $\int_0^3 \frac{3(x+2)}{\sqrt{x+1}} d x$.
A. 32
B. 24
C. 16
D. 18
E. 20
Answer (1)
Use the substitution u = x + 1 to simplify the integral.
Change the limits of integration accordingly: from x = 0 to u = 1 and from x = 3 to u = 4 .
Rewrite and simplify the integral in terms of u : 3 ∫ 1 4 ( u 1/2 + u − 1/2 ) d u .
Evaluate the integral and apply the limits to get the final answer: 20 .
Explanation
Problem Setup We are asked to evaluate the definite integral ∫ 0 3 x + 1 3 ( x + 2 ) d x .
Substitution To solve this integral, we can use a substitution method. Let u = x + 1 . Then, x = u − 1 and d x = d u . Also, we need to change the limits of integration. When x = 0 , u = 0 + 1 = 1 , and when x = 3 , u = 3 + 1 = 4 .
Rewriting the Integral Now we can rewrite the integral in terms of u : ∫ 0 3 x + 1 3 ( x + 2 ) d x = ∫ 1 4 u 3 (( u − 1 ) + 2 ) d u = ∫ 1 4 u 3 ( u + 1 ) d u = 3 ∫ 1 4 u u + 1 d u
Simplifying the Integrand We can simplify the integrand by dividing each term by u : 3 ∫ 1 4 u u + 1 d u = 3 ∫ 1 4 ( u 1/2 + u − 1/2 ) d u
Integrating Now we can integrate term by term: 3 ∫ 1 4 ( u 1/2 + u − 1/2 ) d u = 3 [ 3 2 u 3/2 + 2 u 1/2 ] 1 4
Applying Limits of Integration Now we evaluate the antiderivative at the limits of integration: 3 [ 3 2 u 3/2 + 2 u 1/2 ] 1 4 = 3 [ ( 3 2 ( 4 ) 3/2 + 2 ( 4 ) 1/2 ) − ( 3 2 ( 1 ) 3/2 + 2 ( 1 ) 1/2 ) ]
Simplifying Simplify the expression: 3 [ ( 3 2 ( 8 ) + 2 ( 2 ) ) − ( 3 2 + 2 ) ] = 3 [ ( 3 16 + 4 ) − ( 3 2 + 2 ) ] = 3 [ 3 16 + 4 − 3 2 − 2 ] = 3 [ 3 14 + 2 ]
Final Calculation 3 [ 3 14 + 2 ] = 3 [ 3 14 + 3 6 ] = 3 [ 3 20 ] = 20
Conclusion Thus, the value of the definite integral is 20.
Examples
Definite integrals are used to calculate areas, volumes, and other quantities in physics, engineering, and economics. For example, in physics, the definite integral of a velocity function over a time interval gives the displacement of an object during that time interval. In engineering, definite integrals are used to calculate the work done by a force or the flow rate of a fluid. In economics, they can be used to find the consumer surplus or producer surplus in a market.
