Question Statement Evaluate the following definite integral:
β« 0 1 3 x 4 β 3 x , d x \int_{0}^{1} \frac{3x}{\sqrt{4 - 3x}} , dx β« 0 1 β 4 β 3 x β 3 x β , d x Background and Explanation This integral involves a rational expression with a square root in the denominator. To solve it, we will use substitution to simplify the expression. Understanding the following concepts will help:
Substitution method for simplifying integrals, especially when dealing with expressions like 4 β 3 x \sqrt{4 - 3x} 4 β 3 x β The power rule for integration, especially with powers such as 1 2 \frac{1}{2} 2 1 β
Integration of rational functions and handling square roots in the integrand.
Solution Step 1: Simplify the integrand using substitution
We begin by using substitution to simplify the expression. Letβs use the substitution:
u = 4 β 3 x u = 4 - 3x u = 4 β 3 x Now, differentiate both sides:
d u = β 3 , d x or d x = d u β 3 du = -3 , dx \quad \text{or} \quad dx = \frac{du}{-3} d u = β 3 , d x or d x = β 3 d u β Also, adjust the limits of integration. When x = 0 x = 0 x = 0 u = 4 u = 4 u = 4 x = 1 x = 1 x = 1 u = 1 u = 1 u = 1
Substituting this into the integral:
β« 0 1 3 x 4 β 3 x , d x = β β« 4 1 ( 4 β u ) u β
d u 3 \int_{0}^{1} \frac{3x}{\sqrt{4 - 3x}} , dx = -\int_{4}^{1} \frac{(4 - u)}{\sqrt{u}} \cdot \frac{du}{3} β« 0 1 β 4 β 3 x β 3 x β , d x = β β« 4 1 β u β ( 4 β u ) β β
3 d u β Step 2: Break down the integrand
Simplify the integrand:
β 1 3 β« 4 1 4 β u u , d u -\frac{1}{3} \int_{4}^{1} \frac{4 - u}{\sqrt{u}} , du β 3 1 β β« 4 1 β u β 4 β u β , d u We can break this into two separate integrals:
β 1 3 ( β« 4 1 4 u , d u β β« 4 1 u u , d u ) -\frac{1}{3} \left( \int_{4}^{1} \frac{4}{\sqrt{u}} , du - \int_{4}^{1} \frac{u}{\sqrt{u}} , du \right) β 3 1 β ( β« 4 1 β u β 4 β , d u β β« 4 1 β u β u β , d u ) Step 3: Solve each integral
First integral :
β« 4 1 4 u , d u = 4 β« 4 1 u β 1 2 , d u \int_{4}^{1} \frac{4}{\sqrt{u}} , du = 4 \int_{4}^{1} u^{-\frac{1}{2}} , du β« 4 1 β u β 4 β , d u = 4 β« 4 1 β u β 2 1 β , d u The antiderivative of u β 1 2 u^{-\frac{1}{2}} u β 2 1 β 2 u 1 2 2u^{\frac{1}{2}} 2 u 2 1 β
4 [ 2 u 1 2 ] 4 1 = 8 [ u ] 4 1 = 8 ( 1 β 4 ) = 8 ( 1 β 2 ) = β 8 4 \left[ 2u^{\frac{1}{2}} \right]_{4}^{1} = 8 \left[ \sqrt{u} \right]_{4}^{1} = 8 (\sqrt{1} - \sqrt{4}) = 8 (1 - 2) = -8 4 [ 2 u 2 1 β ] 4 1 β = 8 [ u β ] 4 1 β = 8 ( 1 β β 4 β ) = 8 ( 1 β 2 ) = β 8
Second integral :
β« 4 1 u u , d u = β« 4 1 u 1 2 , d u \int_{4}^{1} \frac{u}{\sqrt{u}} , du = \int_{4}^{1} u^{\frac{1}{2}} , du β« 4 1 β u β u β , d u = β« 4 1 β u 2 1 β , d u The antiderivative of u 1 2 u^{\frac{1}{2}} u 2 1 β 2 3 u 3 2 \frac{2}{3} u^{\frac{3}{2}} 3 2 β u 2 3 β
[ 2 3 u 3 2 ] 4 1 = 2 3 ( ( 1 ) 3 2 β ( 4 ) 3 2 ) = 2 3 ( 1 β 8 ) = 2 3 ( β 7 ) = β 14 3 \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_{4}^{1} = \frac{2}{3} \left( (1)^{\frac{3}{2}} - (4)^{\frac{3}{2}} \right) = \frac{2}{3} \left( 1 - 8 \right) = \frac{2}{3} (-7) = -\frac{14}{3} [ 3 2 β u 2 3 β ] 4 1 β = 3 2 β ( ( 1 ) 2 3 β β ( 4 ) 2 3 β ) = 3 2 β ( 1 β 8 ) = 3 2 β ( β 7 ) = β 3 14 β Step 4: Combine the results
Now, combine the results of both integrals:
β 1 3 ( β 8 β 14 3 ) = 1 3 ( 8 + 14 3 ) -\frac{1}{3} \left( -8 - \frac{14}{3} \right) = \frac{1}{3} \left( 8 + \frac{14}{3} \right) β 3 1 β ( β 8 β 3 14 β ) = 3 1 β ( 8 + 3 14 β ) Convert 8 to a fraction with a denominator of 3:
1 3 ( 24 3 + 14 3 ) = 1 3 β
38 3 = 38 9 \frac{1}{3} \left( \frac{24}{3} + \frac{14}{3} \right) = \frac{1}{3} \cdot \frac{38}{3} = \frac{38}{9} 3 1 β ( 3 24 β + 3 14 β ) = 3 1 β β
3 38 β = 9 38 β Thus, the value of the integral is:
10 9 \boxed{\frac{10}{9}} 9 10 β β Summary of Steps
Substitute u = 4 β 3 x u = 4 - 3x u = 4 β 3 x Rewrite the integrand as two separate integrals:
β« 4 1 4 u , d u β β« 4 1 u u , d u \int_{4}^{1} \frac{4}{\sqrt{u}} , du - \int_{4}^{1} \frac{u}{\sqrt{u}} , du β« 4 1 β u β 4 β , d u β β« 4 1 β u β u β , d u Solve each integral:
First integral: β 8 -8 β 8
Second integral: β 14 3 -\frac{14}{3} β 3 14 β
Combine the results and simplify:
38 9 \frac{38}{9} 9 38 β
The final answer is:
10 9 \boxed{\frac{10}{9}} 9 10 β β 
