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Applications of Integration formulas

11 formulas and 12 common traps for NDA Mathematics Applications of Integration, grouped by subtopic.

Full notes with worked examples

Area Bounded by a Curve, Lines & Axes

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The Definite Integral as Signed Area

Area under a curve above the axis

A=∫abf(x) dx(f≥0)A = \int_a^b f(x)\,dx \quad (f \ge 0)
  • a,ba, bleft and right boundary lines x = a, x = b
  • f(x)f(x)the height of the region at position x

Area Under a Curve Between Two Lines

Semicircle area shortcut

y=r2−x2 ⇒ A=12πr2y = \sqrt{r^2 - x^2}\ \Rightarrow\ A = \tfrac{1}{2}\pi r^2

Below the Axis, Loops & the Factor of 2

Area with a sign change at c

A=∣∫acf dx∣+∣∫cbf dx∣A = \left|\int_a^c f\,dx\right| + \left|\int_c^b f\,dx\right|

Regions Bounded by Lines & Modulus

Polygon area, not an integral

rectangle =w×htriangle =12 b h\text{rectangle } = \text{w}\times\text{h} \qquad \text{triangle } = \tfrac{1}{2}\,b\,h

Area of a Parabola Cut by Its Latus Rectum

Parabola–latus rectum area

y2=4ax:A=2∫0a4ax dx=83a2y^2 = 4ax:\quad A = 2\int_0^{a}\sqrt{4ax}\,dx = \tfrac{8}{3}a^2

Area Under a Step (Greatest-Integer) Curve

One step = one rectangle

A=∣n∣×(interval width),[x]=nA = |n| \times (\text{interval width}), \quad [x] = n

Area of a Circular Segment by a Chord

Segments of a circle

A2=sector−triangle,A1=πr2−A2A_2 = \text{sector} - \text{triangle}, \qquad A_1 = \pi r^2 - A_2

Common traps

Integrate only where the curve stays above the axis

The formula A=∫abf dxA = \int_a^b f\,dx gives the true area only when f≥0f \ge 0 throughout. If the curve crosses the axis inside [a,b][a, b], split the integral at the crossing and take absolute values — see the next concept.

The raw integral can be zero while the area is not

For an odd function over a symmetric interval, ∫−aaf dx=0\int_{-a}^{a} f\,dx = 0. That is the signed integral, not the area. Whenever a region straddles the axis, split and take absolute values — and a symmetric region doubles one half rather than cancelling it.

A negative area answer means a missing modulus

Geometric area is always positive. If a region lies below the axis, ∫abf dx\int_a^b f\,dx comes out negative — that is the SIGNED value, and the area is its magnitude ∣∫abf dx∣\left|\int_a^b f\,dx\right|. Reporting a negative number as 'the area' (forgetting the ∣⋅∣|\cdot|) is the single most common slip in this chapter.

|x| ≤ p gives a side of length 2p, not p

A modulus bound ∣x∣≤p|x| \le p runs from −p-p to +p+p, so the full side is 2p2p. Treating it as length pp halves your dimension and quarters a rectangle's area — the most common modulus-region slip.

Double the half-region, and use the right limit

Two slips combine here: forgetting the factor of 2 (the parabola lies on both sides of the axis), and integrating to x=ax = a the parameter rather than to the actual latus-rectum line. For y2=xy^2 = x the limit is x=14x = \tfrac14, not x=1x = 1.

Negative step values still give positive area

For a negative interval, [x][x] is the lower integer: on [−1.8,−1.5][-1.8, -1.5], [x]=−2[x] = -2 (not −1-1). The rectangle's height is the magnitude ∣−2∣=2|{-2}| = 2. Using −1-1, or letting the area come out negative, are the two traps.

Subtract the triangle from the sector

The minor-segment area is the sector area MINUS the triangle formed by the two radii and the chord — not the whole sector. Computing the segment as the full sector (or as the full integral without removing the triangle) is the standard mistake on these circle-cut questions.

Area Between Two Curves & Intersection Points

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Finding Where Two Curves Meet

Intersection condition

f(x)=g(x) ⇒ the x-values where the curves crossf(x) = g(x) \ \Rightarrow\ \text{the } x\text{-values where the curves cross}

Area Between Curves: Top Minus Bottom

Area between curves

A=∫ab(top−bottom) dxA = \int_a^b \bigl(\text{top} - \text{bottom}\bigr)\,dx

Area Between a Curve and a Line

Curve over a line

A=∫ab(ycurve−yline) dxA = \int_a^b \bigl(y_{\text{curve}} - y_{\text{line}}\bigr)\,dx

Composite Regions: Subtract Areas

Quarter-circle minus a curve

A=14πr2−∫abf(x) dxA = \tfrac{1}{4}\pi r^2 - \int_a^b f(x)\,dx

Common traps

A modulus can create extra intersections

Solving x2=2xx^2 = 2x gives 2 points, but x2=2∣x∣x^2 = 2|x| gives 3 — the modulus mirrors a solution to the negative side. Always account for both signs of ∣x∣|x| when counting crossings.

The intersection x-values are the limits — don't guess them

The limits aa and bb of an area-between-curves integral are the x-coordinates where the curves actually cross, found by solving f(x)=g(x)f(x) = g(x). Using the interval given in the problem text (or the y-intercepts) instead of the true crossings gives the wrong region and the wrong area.

Subtract top minus bottom, not in equation order

The integrand is (upper curve) − (lower curve), decided by which is actually higher between the crossings — NOT the order the curves are named. Subtracting the wrong way gives the negative of the area; if your answer is negative, you reversed them.

Pick the correct branch of a sideways parabola

y2=2xy^2 = 2x has two branches, y=+2xy = +\sqrt{2x} and y=−2xy = -\sqrt{2x}. For a region in the first quadrant against y=xy = x, only the upper branch bounds it — using the full y2y^2 relation without choosing a branch is where the setup breaks.

Subtract the area under the curve, not the curve's value

For 'quarter-circle minus the sine region', you remove ∫0πsin⁡x dx=2\int_0^{\pi}\sin x\,dx = 2, the AREA under sin⁡x\sin x — not a single function value. Mixing up the area with a height (or forgetting the integral evaluates to 2) is the recurring HARD-question slip.

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