Find the area bounded by the lines $y = 4x + 5$, $y = 5 - x$ and $4y = x + 5$.

Find the area bounded by the lines y = 4x + 5, y = – Class 12 Mathematics

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y = 4x + 5

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Concept understanding — Area Under Curve

Area Under a Curve

How do you measure the area of a region whose top edge is curved rather than a straight line? For rectangles and triangles we have formulas, but a shape bounded above by $y = f(x)$ has no simple side lengths to plug in. The definite integral is the tool built exactly for this.

The core idea: area as a limit of strips

Take the region under $y = f(x)$ (with $f(x) \ge 0$ ), above the $x$ -axis, between $x = a$ and $x = b$ . Slice it into many thin vertical strips. A strip at position $x$ with tiny width $dx$ is almost a rectangle of height $f(x)$ , so its area is about $f(x)\, dx$ . Add up all the strips and let their width shrink to zero: the sum becomes the definite integral

\text{Area} = \int_a^b f(x)\, dx.

This is why the integral is the area — it is the exact total of infinitely many infinitesimally thin rectangles.

How to compute it

Find an antiderivative $F(x)$ (so $F'(x) = f(x)$ ) and evaluate at the two limits — the Fundamental Theorem of Calculus:

\int_a^b f(x)\, dx = F(b) - F(a).

For example, the area under $y = x^2$ from $0$ to $1$ is $\left[\tfrac{x^3}{3}\right]_0^1 = \tfrac13$ .

Cases you must handle carefully

Watch out

The integral gives signed area. Where the curve dips below the $x$ -axis, $f(x) < 0$ and $\int f\,dx$ comes out negative. For the geometric (positive) area of such a stretch, integrate the absolute value or take the magnitude of that piece: $\displaystyle \text{Area} = \int_a^b |f(x)|\, dx$ .

If a curve crosses the $x$ -axis inside $[a,b]$ , split the integral at each crossing and add the sizes of the parts.

Area with respect to the y-axis

When the region is bounded by a curve $x = g(y)$ and the $y$ -axis between $y = c$ and $y = d$ , slice horizontally instead: …

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The three lines meet at $(-1,1)$ , $(0,5)$ and $(3,2)$ ; the enclosed triangle has area $\boxed{\dfrac{15}{2}}$ (i.e. $7.5$ ) square units.

Concept

Three non-parallel, non-concurrent lines bound a triangle. Find the three pairwise intersection points, then compute the triangle's area (shoelace formula).

Solution

1. $y=4x+5$ and $y=5-x$ : $4x+5=5-x\Rightarrow 5x=0\Rightarrow x=0,\ y=5$ . Point $A(0,5)$ .

2. $y=4x+5$ and $4y=x+5$ : $4(4x+5)=x+5\Rightarrow 16x+20=x+5\Rightarrow 15x=-15\Rightarrow x=-1,\ y=1$ . Point $B(-1,1)$ .

3. $y=5-x$ and $4y=x+5$ : $4(5-x)=x+5\Rightarrow 20-4x=x+5\Rightarrow 5x=15\Rightarrow x=3,\ y=2$ . Point $C(3,2)$ .

4. Shoelace formula with $A(0,5)$ , $B(-1,1)$ , $C(3,2)$ :

$\text{Area}=\frac12\big|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\big|$ …

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Q.Find the area bounded by the lines $y = 4x + 5$ , $y = 5 - x$ and $4y = x + 5$ .

Concept understanding — Area Under Curve

Area Under a Curve

The core idea: area as a limit of strips

How to compute it

Cases you must handle carefully

Area with respect to the y-axis

Concept

Solution

Method: Area of the triangle cut out by three lines

Steps

Common Mistakes

Q.Find the area bounded by the lines y=4x+5y = 4x + 5y=4x+5, y=5−xy = 5 - xy=5−x and 4y=x+54y = x + 54y=x+5.

Concept

Solution

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Q.Find the area bounded by the lines $y = 4x + 5$ , $y = 5 - x$ and $4y = x + 5$ .

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