Q.Draw a rough sketch of the given curve $y = 1 + |x + 1|$, $x = -3$, $x = 3$, $y = 0$ and find the area of the region bounded by them, using integration.

Concept: Area Under Curve — the region is bounded by a V-shaped absolute-value function and vertical/horizontal lines.

Step 1 – Rewrite the function piecewise

$y = 1 + |x+1|$ means:

• For $x \ge -1$: $y = 1 + (x+1) = x + 2$
• For $x < -1$: $y = 1 - (x+1) = -x$

Step 2 – Sketch the region

The curve has a vertex at $(-1, 1)$. The boundaries are $x = -3$ (left), $x = 3$ (right), and $y = 0$ (x-axis). The region lies above the x-axis between these vertical lines.

Step 3 – Set up the integral

Area splits at $x = -1$:

Step 4 – Evaluate …

Concept

$y=1+|x+1|$ is a V with vertex at $x=-1$. Removing the modulus:

• for $x\ge -1$: $y=1+(x+1)=x+2$;
• for $x<-1$: $y=1-(x+1)=-x$.

The minimum value is $y=1$ at $x=-1$, so the curve lies above $y=0$ throughout $[-3,3]$; the area is just $\int_{-3}^{3}y\,dx$, split at $x=-1$.

Rough sketch

Key points: $(-3,3)$, vertex $(-1,1)$, $(3,5)$. Left arm falls from $(-3,3)$ to $(-1,1)$ (slope $-1$); right arm rises from $(-1,1)$ to $(3,5)$ (slope $+1$). The region is closed below by the $x$-axis and on the sides by $x=-3$ and $x=3$.

Solution

First integral:

Second integral: …

Method: Area under a modulus (piecewise-linear) curve between two ordinates

A curve containing $|\,\cdot\,|$, such as $y = 1 + |x + 1|$, is really two straight lines joined at a corner. To find the area it bounds with the $x$-axis between two vertical lines, break the modulus open and integrate each linear piece.

Steps

Step 1: Rewrite the function without the modulus.

$|x - h|$ equals $(x - h)$ when $x \ge h$ and $-(x - h)$ when $x < h$. This produces two linear formulas, one valid on each side of the corner $x = h$.

Step 2: Check the curve stays above the x-axis.

If the minimum value (reached at the corner) is positive, the whole region lies above $y = 0$, and the area is simply $\int y\,dx$ — no absolute value on the integral is needed. …

Common Mistakes

Mistake 1: Removing the modulus with the wrong sign.

Why it's wrong: $|x+1|$ equals $x+1$ only when $x\ge -1$; for $x<-1$ it is $-(x+1)$. Getting this backwards gives the wrong pieces $y=x+2$ and $y=-x$. Correct approach: split the definition at the vertex $x=-1$.

Mistake 2: Splitting the integral at $x=0$ instead of $x=-1$.

Why it's wrong: the corner of the V — where the formula for $y$ changes — is at $x=-1$, not the origin. Splitting at the wrong place mixes the two arms. Correct approach: integrate $-x$ on $[-3,-1]$ and $x+2$ on $[-1,3]$. …