REFER TO IMAGE ### Calculus Problem: Double Integral Evaluation and Region Sketching#### Problem Statement:4. (a) Evaluate this integral by hand:\[\int_{1}^{3} \int_{4x-3}^{x^2} \frac{6}{x^2} \, dy \, dx\]Show all your steps and give an exact answer.**Solution:***(Step-by-step solution needed here)*---(b) For the integral above, sketch the region of integration, and re-write the integral with the opposite order of integration.**Region of Integration:**To sketch the region of integration, we observe the limits of the integral:- \( x \) ranges from 1 to 3.- For a fixed \( x \), \( y \) ranges from \( 4x-3 \) to \( x^2 \).The region can be visualized on the xy-plane:1. Identify the curves: - \( y = 4x - 3 \) - \( y = x^2 \)2. Mark the points where these curves intersect within the given \( x \)-limits, \( x \in [1, 3] \).**Rewriting the Integral:**To change the order of integration, we need to express the same region in \( y \)-first terms.1. Identify the \( y \)-limits: - The lowest value of \( y \) is when \( x = 1 \) and \( y = 4(1)-3 = 1 \). - The highest value of \( y \) is when \( x = 3 \) and \( y = \min({4(3)-3, 3^2}) = 9.Now rewrite the integral with the new limits:\[\int_{1}^{9} \int_{{\text{left boundary}}}^{\text{{right boundary}}} \frac{6}{x^2} \, dx \, dy\]Here, to correctly determine the new \( x \)-limits as a function of \( y \):- Solve \( y=4x-3 \) for \( x \), which gives \( x = \frac{y+3}{4} \).- Solve \( y = x^2 \) for \( x \), which gives \( x = \sqrt{y} \).Thus, the integral

Name: REFER TO IMAGE ### Calculus Problem: Double Integral Evaluation and Re
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Description: REFER TO IMAGE ### Calculus Problem: Double Integral Evaluation and Region Sketching#### Problem Statement:4. (a) Evaluate this integral by hand:\[\int_{1}^{3} \int_{4x-3}^{x^2} \frac{6}{x^2} \, dy \, dx\]Show all your steps and give an exact answer.