CBSE Class 12 Mathematics sample paper 2025-26: question 28(A)

Hints

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Step-by-step solution

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1. Graph: $y=|x+1|=-x-1$ for $x<-1$ and $x+1$ for $x\ge -1$. It is a V-shape with vertex at $(-1,0)$, passing through $(-4,3)$, $(0,1)$ and $(2,3)$. Draw the lines $x=-4$ and $x=2$ and shade the two triangles between the V and the x-axis from $x=-4$ to $x=2$.

   1 mark

2. ${\int }_{-4}^2|x+1|dx={\int }_{-4}^{-1}(-x-1)dx+{\int }_{-1}^2(x+1)dx$.

   ½ mark

3. $={\left[-\frac{(x+1{)}^2}{2}\right]}_{-4}^{-1}+{\left[\frac{(x+1{)}^2}{2}\right]}_{-1}^2$.

   ½ mark

4. $=-\left(0-\frac{9}{2}\right)+\left(\frac{9}{2}-0\right)=9$.

   ½ mark

5. The value 9 is the area (in sq. units) of the region bounded by $y=|x+1|$, the x-axis and the lines $x=-4$ and $x=2$ (the shaded region).

   ½ mark

Answer: Integral $=9$; it is the area bounded by $y=|x+1|$, the x-axis, $x=-4$ and $x=2$

Revise the chapter

• Chapter: Application of Integrals (notes, formulas, where marks are lost and every past question)