NDA (National Defence Academy) · Mathematics (JEE & NDA)
Integral Calculus
Indefinite integration by substitution, parts and partial fractions, definite integrals, and areas bounded by curves.
Seven concepts. Indefinite techniques first, then the definite-integral properties JEE actually scores — king's rule, even/odd limits, and area taken positive.
- NDA (National Defence Academy)
- Hard level
- 7 concepts
- 5 practice questions
1Indefinite integral and standard forms
Integration reverses differentiation: \int f'(x)\,dx = f(x)+C. The constant of integration is mandatory — every antiderivative differs by a constant, and dropping it loses a mark. The power rule \int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C holds only for n\ne -1; the missing case is \int\dfrac{dx}{x}=\ln|x|+C.
A three-row sheet: power (with n\ne -1), logarithm, and exponential — each written as integrand beside antiderivative plus C.
Check an antiderivative
- Write a candidateGuess F from a standard form, keeping +C.
- DifferentiateCompute F'(x) and confirm it equals the integrand.
- Watch n=-1If the integrand is 1/x, use \ln|x|, never the power rule.
| Integrand | Antiderivative |
|---|---|
| x^n (n\ne -1) | \dfrac{x^{n+1}}{n+1}+C |
| 1/x | \ln|x|+C |
| e^x | e^x+C |
Which antiderivative is correct for \int x^{-1}\,dx?
- \dfrac{x^{0}}{0}+C
- \ln|x|+C
- x^{-1}+C
The power rule excludes n=-1; that case is the logarithm. The first option is undefined, and x^{-1} is the integrand, not an antiderivative.
2Substitution
Substitution undoes the chain rule: \int f(g(x))g'(x)\,dx becomes \int f(u)\,du after u=g(x). The whole point is to make the factor g'(x) absorb into du, so choose the inner function whose derivative is already visible (or nearly visible) in the integrand. Rational functions that stay rational after a linear substitution are then split by partial fractions.
Figure. In (2x+1)e^{x^2+x}\,dx the inner composite is u=x^2+x, whose differential is already sitting in front as du=(2x+1)\,dx. The integrand becomes e^u\,du.
Running a substitution
- Name uSet u equal to the inner composite that sits under a power, root, or trig function.
- Match duDifferentiate: du=g'(x)\,dx. Rearrange the integrand until that factor appears.
- Integrate in uEvaluate \int f(u)\,du, then replace u by g(x) and add C.
- Partial fractions if neededWhen the result in u is a proper rational function, factor the denominator and split before integrating.
In \int (2x+1)\,e^{x^2+x}\,dx, the useful substitution starts with
- u=e^{x}
- u=x^2+x
- u=2x+1
u=x^2+x has du=(2x+1)\,dx, which is already the remaining factor. Taking u=2x+1 leaves the exponential unmatched; u=e^{x} ignores the quadratic exponent.
3Integration by parts
The product rule backwards is \int u\,dv = uv - \int v\,du. Choose u so that differentiating it simplifies, and dv so that integrating it is still possible. The ILATE order — Inverse, Logarithmic, Algebraic, Trigonometric, Exponential — picks which factor becomes u: the earlier category is differentiated.
Figure. ILATE takes algebraic u=x and leaves dv=e^{x}\,dx. Then du=dx, v=e^{x}, so \int xe^{x}\,dx=xe^{x}-\int e^{x}\,dx=e^{x}(x-1)+C.
ILATE choice
- Rank the factorsPlace each factor of the product on the ILATE ladder.
- Set uThe factor from the earlier category is u; the rest is dv.
- Form du and vDifferentiate u, integrate dv, then write uv-\int v\,du.
Integration by parts
Evaluate \displaystyle\int x e^{x}\,dx.
- ILATE: u=x (algebraic), dv=e^{x}\,dxdu=dx, v=e^{x}
- \int xe^{x}\,dx = uv - \int v\,duxe^{x} - \int e^{x}\,dx
- xe^{x} - e^{x}+Ce^{x}(x-1)+C
Pro tip. By ILATE the algebraic factor x is chosen as u; differentiating it to a constant ends the process after one step.
For \int x\ln x\,dx, ILATE says take u=
- x (algebraic before logarithmic)
- \ln x (logarithmic before algebraic)
- x\ln x as a single factor
Logarithmic precedes Algebraic on ILATE, so u=\ln x and dv=x\,dx. Choosing u=x forces integrating \ln x, which is the harder factor.
4Fundamental theorem and signed area
The fundamental theorem of calculus evaluates a definite integral from any antiderivative: \int_a^b f(x)\,dx = F(b)-F(a) whenever F'=f. Geometrically that number is the signed area between y=f(x) and the x-axis on [a,b] — regions below the axis subtract.
Figure. On [a,b] the definite integral equals the signed area under y=f(x). Here f(x)=x stays above the axis, so the sign is positive; a dip below would subtract.
Evaluating a definite integral
- Find FProduce an antiderivative of f by a standard form or a technique from the earlier concepts.
- Subtract endsCompute F(b)-F(a). The +C cancels, so it may be omitted here.
- Read the signA negative result means more area lies below the axis than above — the integral is signed, not a geometric area.
If F'=f and \int_0^1 f(x)\,dx = 3, then F(1)-F(0) equals
- 3
- F(1)+F(0)
- an unknown constant C
By the FTC the definite integral is exactly F(b)-F(a). Adding the ends or reintroducing C confuses the indefinite form with the definite evaluation.
5King's rule
King's rule replaces the dummy variable without changing the value: \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx. On [0,\pi/2] the map is x\mapsto \tfrac{\pi}{2}-x, which swaps sine and cosine. Adding the original integral to its image often cancels a hard quotient and leaves a constant integrand.
Figure. King's rule replaces x by \pi/2-x and swaps sine with cosine. Adding the two forms of I=\int_0^{\pi/2}\cos x/(\sin x+\cos x)\,dx cancels the denominator and leaves 2I=\pi/2, so I=\pi/4.
Using the partner integral
- Write IName the definite integral you want.
- Replace xApply x\to a+b-x to obtain a second expression for the same I.
- Add the copiesAdd the two forms. If the integrands sum to a constant, finish in one line.
Definite integral via king's rule
Evaluate I=\displaystyle\int_0^{\pi/2}\dfrac{\sin x}{\sin x+\cos x}\,dx.
- x\to\tfrac{\pi}{2}-xI=\displaystyle\int_0^{\pi/2}\dfrac{\cos x}{\cos x+\sin x}\,dx
- Add both forms: 2I=\displaystyle\int_0^{\pi/2}\dfrac{\sin x+\cos x}{\sin x+\cos x}\,dx2I=\int_0^{\pi/2}1\,dx=\dfrac{\pi}{2}
- I=\dfrac{\pi}{4}\dfrac{\pi}{4}
Pro tip. When the integrand plus its x\to a+b-x partner simplifies to a constant, adding the two copies solves the integral in one line. Before integrating \int_0^{\pi/2} trig powers, try this replacement first.
If I=\int_0^{\pi/2}\dfrac{\cos x}{\sin x+\cos x}\,dx, then I equals
- \dfrac{\pi}{2}
- \dfrac{\pi}{4}
- 1
This I is exactly the partner of \int_0^{\pi/2}\sin x/(\sin x+\cos x)\,dx under king's rule, so the same addition gives I=\pi/4.
6Even and odd integrands
Symmetric limits turn parity into an evaluation shortcut: \int_{-a}^{a}f(x)\,dx = 2\int_0^{a}f(x)\,dx when f is even, and equals 0 when f is odd (and the integral exists). Spot an odd integrand on a symmetric interval and write zero instead of hunting an antiderivative.
Figure. An odd integrand is rotationally symmetric about the origin: the lobe in the third quadrant cancels the lobe in the first, so the integral over symmetric limits is zero.
| Parity | Test | Definite integral |
|---|---|---|
| Even | f(-x)=f(x) | 2\int_0^{a}f(x)\,dx |
| Odd | f(-x)=-f(x) | 0 |
\int_{-2}^{2} x^3\,dx equals
- 8
- 0
- 16
x^3 is odd and the limits are symmetric, so the integral is 0. The positive values 8 and 16 are what you get if you integrate only on [0,2] or forget the sign cancellation.
7Area between curves
Geometric area is never negative. The area between y=f(x) and y=g(x) from x=a to x=b is A=\int_a^b |f(x)-g(x)|\,dx — equivalently, integrate higher curve minus lower curve on each sub-interval where which one is higher stays fixed. A raw \int(f-g) without the absolute value returns signed net area, which is the wrong object when the question asks for area.
Figure. Schematic only: solid y=x above dashed y=x^2 on (0,1). Area is \int_0^1(x-x^2)\,dx — higher minus lower — not the signed integral of either curve alone.
Setting up the area integral
- Find intersectionsSolve f(x)=g(x) for the limits (or the split points) of the integral.
- Name higher and lowerOn each interval, identify which graph sits above the other.
- Integrate the differenceWrite \int(\text{higher}-\text{lower})\,dx, or use |f-g| in one piece when that is cleaner.
Area between two curves on [a,b] is computed as
- \int_a^b (f-g)\,dx with the sign kept
- \int_a^b |f-g|\,dx
- \bigl|\int_a^b (f-g)\,dx\bigr| always
Absolute value goes inside so each strip is positive. Taking the absolute value of the whole integral after integrating can cancel opposing lobes and under-count geometric area.
Notes
- Indefinite integration: Integration reverses differentiation, \int f'(x)\,dx=f(x)+C. Standard results include \int x^n dx=\dfrac{x^{n+1}}{n+1} (n\ne-1), \int\dfrac{dx}{x}=\ln|x| and \int e^x dx=e^x.
- Techniques: Substitution converts \int f(g(x))g'(x)dx into \int f(u)du. Integration by parts uses \int u\,dv=uv-\int v\,du (choose u by the ILATE order). Rational functions are split by partial fractions.
- Fundamental theorem of calculus: \int_a^b f(x)\,dx=F(b)-F(a), where F'=f. The definite integral gives the signed area between the curve and the x-axis.
- Properties of definite integrals: \int_a^b f(x)dx=\int_a^b f(a+b-x)dx (king's rule) and \int_{-a}^{a}f(x)dx=2\int_0^a f(x)dx if f is even, or 0 if f is odd. These often bypass a difficult antiderivative.
- Area under and between curves: The area between y=f(x) and y=g(x) from a to b is \int_a^b|f(x)-g(x)|\,dx; areas are always taken positive by using the higher curve minus the lower one.
Formulas
- Power / log: \int x^n dx=\dfrac{x^{n+1}}{n+1}+C,\quad \int\dfrac{dx}{x}=\ln|x|+C
- By parts: \int u\,dv=uv-\int v\,du
- Fundamental theorem: \int_a^b f(x)dx=F(b)-F(a)
- King's rule: \int_a^b f(x)dx=\int_a^b f(a+b-x)dx
- Symmetry: \int_{-a}^{a}f\,dx=2\int_0^a f\,dx (even), 0 (odd)
- Area between curves: A=\int_a^b|f(x)-g(x)|\,dx
Exam traps & shortcuts
- Choose u in integration by parts using ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) - the earlier category is differentiated.
- Before integrating \int_0^{\pi/2} trig powers, try the king's rule replacement x\to\tfrac{\pi}{2}-x; adding the two forms often gives a trivial integral.
- Spot odd integrands over symmetric limits and write the answer as zero instantly instead of finding an antiderivative.
Reference tables
The six formulas this topic leans on, collected for revision.
| Name | Statement |
|---|---|
| Power / log | \int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C (n\ne -1); \int\dfrac{dx}{x}=\ln|x|+C |
| By parts | \int u\,dv = uv - \int v\,du |
| Fundamental theorem | \int_a^b f(x)\,dx = F(b)-F(a) |
| King's rule | \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx |
| Even / odd | \int_{-a}^{a}f=2\int_0^{a}f (even); 0 (odd) |
| Area between curves | A=\int_a^b |f(x)-g(x)|\,dx |
Recap
Read only this the night before.
- +C and n = −1
- Indefinite integrals keep +C. Power rule skips n=-1; that case is \ln|x|.
- Substitution
- Set u to the inner function whose derivative is already in the integrand; finish in u, then back-substitute.
- Parts / ILATE
- Earlier ILATE category is u. Algebraic x against e^{x} gives e^{x}(x-1)+C in one step.
- FTC
- \int_a^b f = F(b)-F(a) is signed area — below the axis subtracts.
- King's rule
- Add f(x) to f(a+b-x). On [0,\pi/2], \sin/(\sin+\cos) becomes \pi/4.
- Odd on [−a, a]
- Odd integrand, symmetric limits → write 0 immediately.
- Area
- Higher minus lower, or |f-g| inside the integral — geometric area stays positive.
Practise Integral Calculus
Reading is free and needs no account. Practice, mocks and progress live in the app.
- 5 exam-style questions on this topic, with explanations
- A 5-question practice set that ends the chapter
- Timed mocks scored with the real marking scheme
- Readiness tracked per topic, kept on your device