Evaluate the Integral

Problem

(∫_0^3)(|2*x−3|*d(x))

Solution

1. Identify the critical point where the expression inside the absolute value changes sign by setting 2*x−3=0

2*x−3=0

x=3/2

2. Split the integral into two parts based on the critical point x=1.5 which lies within the interval [0,3]

(∫_0^3)(|2*x−3|*d(x))=(∫_0^3/2)(−(2*x−3)*d(x))+(∫_3/2^3)((2*x−3)*d(x))

3. Simplify the integrands by distributing the negative sign in the first integral.

(∫_0^3/2)((3−2*x)*d(x))+(∫_3/2^3)((2*x−3)*d(x))

4. Find the antiderivatives for both expressions using the power rule.

(∫_^)((3−2*x)*d(x))=3*x−x2

(∫_^)((2*x−3)*d(x))=x2−3*x

5. Apply the Fundamental Theorem of Calculus by evaluating the antiderivatives at the upper and lower limits for each part.

[3*x−x2](3/2)0=(3*(3/2)−(3/2)2)−(0)

[x2−3*x]3(3/2)=(3−3*(3))−((3/2)2−3*(3/2))

6. Calculate the values for each definite integral.

9/2−9/4=9/4

0−(9/4−9/2)=9/4

7. Sum the results of the two parts to find the total area.

9/4+9/4=18/4

18/4=9/2

Final Answer

(∫_0^3)(|2*x−3|*d(x))=9/2

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