Chapter 8 · Page 4 of 5

Integration
with Graphs

Sketching standard curves, identifying bounded regions, and solving area problems graphically.

Class 12 · Mathematics · CBSE & JEE
01 · Why Draw a Graph?
The Role of Sketching in Area Problems

A rough sketch is not optional — it is the most important step. Without a sketch you cannot:

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CBSE Mark Allocation

In a 5-mark area problem: 1 mark is typically given just for a correct sketch with the shaded region. Never skip it!

02 · Sketching Guide
How to Sketch Standard Curves Quickly
y = x²

Upward Parabola

  • Vertex at (0, 0)
  • Passes through (1,1), (−1,1), (2,4)
  • Symmetric about y-axis
y² = 4ax

Rightward Parabola

  • Vertex at (0, 0)
  • Opens right; passes (a, 2a)
  • Symmetric about x-axis
x² + y² = r²

Circle

  • Centre (0,0), radius r
  • Mark (\(r\),0), (0,\(r\)), etc.
  • Use 4-fold symmetry
x²/a² + y²/b² = 1

Ellipse

  • Vertices: (±a, 0), (0, ±b)
  • Major axis along x if a > b
  • 4-fold symmetric
y = sin x / cos x

Trig Curves

  • Mark zeros and maxima/minima
  • sin: 0 at nπ, max at π/2
  • cos: 0 at (2n+1)π/2
y = mx + c

Straight Line

  • Find x-intercept: y=0
  • Find y-intercept: x=0
  • Join the two points
03 · Identifying Region
How to Identify the Bounded Region

After sketching, follow this process to correctly identify the shaded (required) region:

Find all intersection points of the given curves/lines. Mark them on the sketch.

Shade the enclosed region — the area completely surrounded by the given boundaries.

Determine limits of integration — the x-coordinates (or y-coordinates) of the intersection points become your limits.

Choose integration variable — if boundaries are easier to express as \(x = g(y)\), integrate w.r.t. y.

Check sign — substitute a test point from inside the region to confirm which function is upper/right boundary.

04 · Visual Example
Bounded Region — Parabola \(y=x^2\) and Line \(y=x+2\)

Graph: y = x² (blue) and y = x + 2 (gold) — Shaded Region

x y 1 2 3 -1 -2 1 2 3 -1 (-1,1) (2,4) y = x² y = x+2 Area
Solution using the graph
Area enclosed by \(y = x^2\) and \(y = x + 2\)

From graph: Intersection points are \((-1, 1)\) and \((2, 4)\). The line \(y = x+2\) is above the parabola in this interval.

\(A = \displaystyle\int_{-1}^2 \big[(x+2) - x^2\big]\,dx = \left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^2\)

\(= \left(2+4-\dfrac{8}{3}\right)-\left(\dfrac{1}{2}-2+\dfrac{1}{3}\right) = \dfrac{10}{3}-\left(-\dfrac{7}{6}\right) = \dfrac{20}{6}+\dfrac{7}{6} = \mathbf{\dfrac{9}{2}}\) sq. units

05 · Integration w.r.t. y
When to Integrate with Respect to y

When the region is naturally described by horizontal strips, integrating w.r.t. \(y\) is much cleaner. Use this when curves are given as \(x = g(y)\).

Area via horizontal strips
\[ A = \int_c^d \big[x_{\text{right}}(y) - x_{\text{left}}(y)\big]\,dy \]
Example — w.r.t. y
Area bounded by \(y^2 = x\) and \(x = y + 2\).

Rewrite: \(x_{\text{right}} = y + 2\), \(x_{\text{left}} = y^2\).

Intersections: \(y^2 = y+2 \Rightarrow y^2-y-2=0 \Rightarrow (y-2)(y+1)=0 \Rightarrow y=-1,\;2\).

\(A = \displaystyle\int_{-1}^2 [(y+2)-y^2]\,dy = \left[\dfrac{y^2}{2}+2y-\dfrac{y^3}{3}\right]_{-1}^2 = \mathbf{\dfrac{9}{2}}\) sq. units

Notice: same numerical answer as previous example — coincidence! But the method is different.

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Board Exam — Graph Tips

  • For a circle, mark four points: (r,0), (0,r), (-r,0), (0,-r).
  • For parabola \(y^2=4ax\): mark focus \((a,0)\) and latus rectum at \(x=a\).
  • Always label the intersection points with coordinates on the graph.
  • Shade the required region and write "Required Area" with an arrow.
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JEE — Graph-Based Questions

  • JEE often gives a graph and asks for the area directly — read intersection points from the graph carefully.
  • Parametric curves: sketch using \((x(t), y(t))\) at key values of \(t\).
  • For \(y = f(|x|)\): use symmetry about y-axis — sketch for \(x \geq 0\) then reflect.
  • Practice recognising regions quickly: parabola-line, two parabolas, circle-line.