Integration as the inverse machine
Integration is the reverse of differentiation. If a function \(F(x)\) has derivative \(f(x)\) — that is, \(\dfrac{dF}{dx}=f(x)\) — then \(F\) is called an antiderivative of \(f\), and the operation that recovers \(F\) from \(f\) is written \(\int f(x)\,dx = F(x)+C\). The elongated S of the integral sign is a stretched "sum"; the \(dx\) names the variable you are integrating over. Where differentiation answers "how fast is this changing?", integration answers "given the rate of change, what is the accumulated total?"
This inverse relationship is the practical engine of the subject. Every entry in the table of derivatives, read from right to left, becomes an entry in the table of integrals. Because \(\dfrac{d}{dx}\!\left(\dfrac{x^{n+1}}{n+1}\right)=x^{n}\), the integral of \(x^n\) must be \(\dfrac{x^{n+1}}{n+1}\). Because \(\dfrac{d}{dx}(\sin x)=\cos x\), the integral of \(\cos x\) is \(\sin x\). You do not memorise integration rules separately; you memorise the derivatives and run them backwards.
There is, however, a second and equally important picture: integration as the area under a curve. Slice the region between a graph \(y=f(x)\) and the \(x\)-axis into thin vertical strips of width \(dx\) and height \(f(x)\); each strip has area \(f(x)\,dx\), and the integral adds them up. The remarkable fact — the fundamental theorem of calculus — is that this geometric sum equals the antiderivative evaluated at the endpoints. The two faces are one.
If \(\dfrac{dF}{dx}=f(x)\), then \(\displaystyle\int f(x)\,dx=F(x)+C\) and \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\). The first is a family of functions; the second is a single number — the signed area under \(f\) from \(a\) to \(b\).
The indefinite integral and the +C
An indefinite integral \(\int f(x)\,dx\) is not a single function but a whole family, all differing by a constant. The reason lies in differentiation itself: the derivative of any constant is zero, so \(x^2\), \(x^2+3\) and \(x^2-7\) all have the same derivative \(2x\). Running the process backwards, you cannot tell which constant was present — so every antiderivative of \(2x\) is written \(x^2+C\), with \(C\) standing for that lost, unknowable constant.
Geometrically the \(+C\) is a vertical shift. The family \(x^2+C\) is an infinite stack of identical parabolas, each one slid up or down. They all have the same slope \(2x\) at every value of \(x\), which is exactly why they share a derivative. In physics the constant is rarely left dangling: an initial condition nails it down. Knowing that a particle was at position \(x_0\) when \(t=0\), or that it started from rest, supplies one equation that fixes \(C\) to a single value.
Standard integrals to memorise
These seven results cover almost every integral that NEET physics demands. Read each one as a derivative reversed; the right-hand "why" column shows which derivative it undoes.
| Integral | Result | Why (derivative reversed) |
|---|---|---|
∫ xⁿ dx | \(\dfrac{x^{n+1}}{n+1}+C\) | Holds for \(n\neq -1\). The "raise the power by one, divide by the new power" rule. |
∫ (1/x) dx | \(\ln|x|+C\) | The exception to the power rule, the case \(n=-1\) where the formula above would divide by zero. |
∫ sin x dx | \(-\cos x+C\) | Reverse of \(\dfrac{d}{dx}(\cos x)=-\sin x\); the minus sign migrates here. |
∫ cos x dx | \(\sin x+C\) | Reverse of \(\dfrac{d}{dx}(\sin x)=\cos x\); no sign change. |
∫ eˣ dx | \(e^{x}+C\) | The function that is its own derivative is also its own integral. |
∫ dx | \(x+C\) | The \(n=0\) case of the power rule; the area under a constant height 1. |
∫ e^{-λt} dt | \(-\dfrac{1}{\lambda}e^{-\lambda t}+C\) | Chain rule in reverse; appears in capacitor discharge and radioactive decay. |
Two linearity rules ride alongside the table and are used in almost every problem. A constant slides outside the integral, \(\int k\,f(x)\,dx = k\int f(x)\,dx\); and an integral of a sum splits into a sum of integrals, \(\int\!\big(f+g\big)\,dx=\int f\,dx+\int g\,dx\). Together with the table they let you integrate any polynomial term by term — which is the bulk of NEET kinematics.
The definite integral and the area under a curve
A definite integral \(\int_a^b f(x)\,dx\) carries two limits — a lower limit \(a\) and an upper limit \(b\) — and evaluates to a single number, not a function. By the fundamental theorem of calculus it equals \(F(b)-F(a)\), where \(F\) is any antiderivative of \(f\). The constant \(C\) cancels in the subtraction: \(\big(F(b)+C\big)-\big(F(a)+C\big)=F(b)-F(a)\). That is why a definite integral never carries a \(+C\), and why any antiderivative will do.
Its geometric meaning is the signed area between the curve \(y=f(x)\) and the horizontal axis, from \(x=a\) to \(x=b\). "Signed" because area lying above the axis counts positive and area below counts negative; an integral over a region equally split above and below can be zero. This single idea — area under the right curve — is what makes integration the language of accumulated physical quantities.
What integration means in physics
The reason integration pervades mechanics is that physical totals are accumulated rates. Velocity is the rate of change of position, so position must be the accumulation — the integral — of velocity. The same logic chains upward and outward across the syllabus, giving four integrals worth holding in active memory.
| Physical quantity | As an integral | Area under which graph |
|---|---|---|
| Velocity from acceleration | \(v=\displaystyle\int a\,dt\) | Area under the acceleration–time \((a\!-\!t)\) graph |
| Displacement from velocity | \(x=\displaystyle\int v\,dt\) | Area under the velocity–time \((v\!-\!t)\) graph |
| Work from a variable force | \(W=\displaystyle\int F\,dx\) | Area under the force–displacement \((F\!-\!x)\) graph |
| Impulse from a force | \(J=\displaystyle\int F\,dt\) | Area under the force–time \((F\!-\!t)\) graph |
Each row is the same statement in a different costume: a quantity equals the area under the graph of its rate. The definite versions read off real intervals — displacement between two instants, work between two positions, impulse during a collision. The constant of integration vanishes the moment you make the integral definite, which is why these physical quantities come out as clean numbers.
Worked example 1 — velocity and position from acceleration
A particle starts from rest at the origin and moves along a straight line with acceleration \(a(t)=6t~\text{m s}^{-2}\). Find its velocity and position as functions of time.
Velocity. Integrate the acceleration: \(v=\int a\,dt=\int 6t\,dt = 6\cdot\dfrac{t^{2}}{2}+C_1 = 3t^{2}+C_1\). The particle starts from rest, so \(v(0)=0\Rightarrow C_1=0\). Hence \(v(t)=3t^{2}~\text{m s}^{-1}\).
Position. Integrate the velocity: \(x=\int v\,dt=\int 3t^{2}\,dt = 3\cdot\dfrac{t^{3}}{3}+C_2 = t^{3}+C_2\). It starts at the origin, so \(x(0)=0\Rightarrow C_2=0\). Hence \(x(t)=t^{3}~\text{m}\).
Check. Differentiating back, \(\dfrac{dx}{dt}=3t^{2}=v\) and \(\dfrac{dv}{dt}=6t=a\). The two integration constants were each fixed by one initial condition — exactly as the \(+C\) demands.
Notice the workflow: integrate once for \(v\), fix \(C_1\) with the initial velocity; integrate again for \(x\), fix \(C_2\) with the initial position. For the special case of constant acceleration this same double integration regenerates \(v=u+at\) and \(x=ut+\tfrac12at^2\) — the familiar kinematic equations are simply integrals in disguise.
Worked example 2 — displacement as the area under a v–t graph
A body moves with velocity \(v(t)=4+2t~\text{m s}^{-1}\). Find its displacement between \(t=1~\text{s}\) and \(t=3~\text{s}\), and confirm it equals the area under the velocity–time graph.
By definite integral. \(x=\displaystyle\int_{1}^{3}(4+2t)\,dt = \Big[\,4t+t^{2}\,\Big]_{1}^{3}\). Upper limit: \(4(3)+3^{2}=12+9=21\). Lower limit: \(4(1)+1^{2}=4+1=5\). Displacement \(=21-5=16~\text{m}\).
By area. The \(v\!-\!t\) graph is a straight line; between \(t=1\) and \(t=3\) it forms a trapezium with parallel sides \(v(1)=6\) and \(v(3)=10\), and width \(2~\text{s}\). Area \(=\tfrac12(6+10)\times 2 = 16~\text{m}\) — identical.
Reading. The two methods must agree: the definite integral is the area under the curve. For straight-line graphs you can use the geometric area as a shortcut; for curved graphs the integral is the only route.
Worked example 3 — work done by a spring
A spring of stiffness \(k\) exerts a restoring force \(F(x)=-kx\). Find the work done by the spring as the block moves from the natural length \(x=0\) to an extension \(x=A\).
Why an integral. The force is not constant — it grows with extension — so \(W=F\cdot d\) does not apply. The correct definition is \(W=\displaystyle\int_{0}^{A}F(x)\,dx\).
Evaluate. \(W=\displaystyle\int_{0}^{A}(-kx)\,dx = -k\Big[\dfrac{x^{2}}{2}\Big]_{0}^{A} = -k\left(\dfrac{A^{2}}{2}-0\right) = -\dfrac{1}{2}kA^{2}\).
Interpretation. The work done by the spring is negative as it is stretched, because the restoring force opposes the displacement. The work done against the spring — stored as elastic potential energy — is \(+\tfrac12 kA^{2}\). Geometrically this is the area of the triangle under the \(F\!-\!x\) line, with base \(A\) and height \(kA\): area \(=\tfrac12(A)(kA)=\tfrac12 kA^{2}\).
This is the cleanest demonstration of why physics needs the integral and not the product. A constant force does work \(F\,d\), the area of a rectangle. A variable force does work equal to the area under its \(F\!-\!x\) graph, which the integral computes exactly. The spring result \(\tfrac12 kA^2\) is the foundation of every NEET problem on elastic potential energy and simple harmonic motion.
Integration in one breath
- Integration reverses differentiation: if \(\frac{dF}{dx}=f\), then \(\int f\,dx=F+C\). Read the derivatives table backwards.
- The indefinite integral is a family of functions; always write \(+C\), then fix it with an initial condition.
- Power rule \(\int x^n dx=\frac{x^{n+1}}{n+1}+C\) for every \(n\) except \(-1\); the exception is \(\int\frac{dx}{x}=\ln|x|+C\).
- The definite integral \(\int_a^b f\,dx = F(b)-F(a)\) is a single number — the signed area under the curve; no \(+C\).
- Subtract lower-limit from upper-limit value; swapping the limits flips the sign.
- Physics integrals: \(v=\int a\,dt\), \(x=\int v\,dt\), \(W=\int F\,dx\), \(J=\int F\,dt\) — each a "quantity = area under its rate".