Q.Find the area of the region bounded by the triangle whose vertices are $(-1, 1)$, $(0, 5)$ and $(3, 2)$, using integration.

Concept: Area under a curve — the area of a triangle can be found by integrating the difference between the upper and lower boundary lines over the appropriate $x$-interval.

Step 1 – Equations of the sides

• Side $AB$ (from $(-1,1)$ to $(0,5)$): slope $= \frac{5-1}{0+1}=4$, equation $y = 4x + 5$.
• Side $BC$ (from $(0,5)$ to $(3,2)$): slope $= \frac{2-5}{3-0} = -1$, equation $y = -x + 5$.
• Side $AC$ (from $(-1,1)$ to $(3,2)$): slope $= \frac{2-1}{3+1} = \frac14$, equation $y = \frac14 x + \frac54$.

Step 2 – Set up the integrals

The region is split at $x=0$ because the upper boundary changes.

For $-1 \le x \le 0$: upper line is $AB$ ($4x+5$), lower line is $AC$ ($\frac14 x + \frac54$).

For $0 \le x \le 3$: upper line is $BC$ ($-x+5$), lower line is $AC$ ($\frac14 x + \frac54$).

Step 3 – Compute

Simplify each integrand:

First: $4x+5 - \frac14 x - \frac54 = \frac{15}{4}x + \frac{15}{4} = \frac{15}{4}(x+1)$. …

Concept

The area enclosed by the three sides equals $\int(\text{upper boundary}-\text{lower boundary})\,dx$ over the $x$-span. The upper edge switches at the middle vertex, so the integral is split there; the lower edge is a single line throughout.

Solution

1. Equations of the sides (two-point form) for $A(-1,1)$, $B(0,5)$, $C(3,2)$:

• $AB$: slope $\dfrac{5-1}{0-(-1)}=4\Rightarrow y=4x+5$
• $BC$: slope $\dfrac{2-5}{3-0}=-1\Rightarrow y=-x+5$
• $AC$: slope $\dfrac{2-1}{3-(-1)}=\dfrac14\Rightarrow y=\dfrac{x}{4}+\dfrac54$

2. Boundaries. $AC$ is the lower edge throughout (at $x=0$, $AC$ gives $1.25$ vs $AB,BC$ giving $5$). The upper edge is $AB$ on $[-1,0]$ and $BC$ on $[0,3]$.

3. Set up the integrals.

Simplify the integrands:

4. Evaluate. …

Method: Area of a triangle by integration (split at the middle vertex)

This technique finds the area of a triangle from its vertices using definite integrals rather than a ready-made formula, exactly as an "using integration" question demands.

Steps

Step 1: Find the equations of the three sides.

From the vertices, use the two-point form to get each side as a line $y = mx + c$. You will have three such lines.

Step 2: Identify the upper and lower boundaries.

One side runs along the bottom of the triangle for the whole $x$-span; the other two form the top but switch at the middle vertex. Sort the vertices by their $x$-coordinates so you know where that switch occurs.

Step 3: Split the integral at the middle vertex's $x$-coordinate. …

Common Mistakes

Mistake 1: Not splitting the integral at the middle vertex $x = 0$

Why it's wrong: the upper boundary is side $AB$ ($y = 4x + 5$) on $[-1, 0]$ but switches to side $BC$ ($y = -x + 5$) on $[0, 3]$; using one line for the whole span mis-measures the triangle. Correct approach: integrate (upper $-$ lower) separately over $[-1,0]$ and $[0,3]$ and add.

Mistake 2: Misidentifying the lower boundary

Why it's wrong: side $AC$ ($y = \tfrac{x}{4} + \tfrac54$) is the lower edge across the whole base (at $x = 0$ it gives $1.25$, below the top value $5$); swapping it with a top side flips signs. Correct approach: subtract $AC$ from whichever upper side applies on each subinterval. …