Using integration, find the area of the region bounded by the line y−1=x, the x-axis, and x=−2 and x=3.

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The given equations are :y−1=x    ..... ( 1 )x=−2    ...... ( 2 ) x=3        ..... ( 3 )Equation (1)  represents a straight line intersecting x−axis at (−1,0) and y−axis at (0,1).Equation (2) represents a straight line parallel to y−axis at a distance of 2 units on the left of y−axis.Equation (3) represents a straight line parallel to y−axis at a distance of 3 units on the right of y−axis.Points of intersection:put x=−2 in (1), we get y=−1put x=3 in (1), we get y=4Let AB be the given line, intersecting the x−axis at C(−1,0).Required area = Area CDAC+ Area CBEC                         =∫−13​ydx+∫−2−1​(−y)dx                         =∫−13​(x+1)dx+∫−2−1​−(x+1)dx                         =[2x2​+x]−13​−[2x2​+x]−2−1​                         =[(29​+3)−(21​−1)]−[(21​−1)−(2−2)]                         =215​+21​+21​                         =217​                         =8.5 square unit

Question Text	Using integration, find the area of the region bounded by the line $y - 1 = x$ , the x-axis, and $x = - 2$ and $x = 3$ .
Answer Type	Text solution:1
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Using integration, find the area of the region bounded by the line $y - 1 = x$ , the x-axis, and $x = - 2$ and $x = 3$ .

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