Q.Evaluate ∫−11ex+e−xexdx
Concept understanding — Polynomial Integration
Polynomial Integration — From Intuition to Precision
Think of integration as the reverse of differentiation. If differentiation tells you the slope of a curve at every point, integration tells you the area under that curve between two points. For polynomials, this reverse process is beautifully simple.
The Intuition: Undoing the Power Rule
You already know the power rule for differentiation: if f(x)=xn, then f′(x)=nxn−1. Integration asks: given that the derivative is xn, what was the original function?
Suppose you want a function whose derivative is x2. You need something that, when differentiated, gives x2. Try x3: its derivative is 3x2, which is three times too big. So try 31x3 — its derivative is exactly x2. That's the core idea: increase the exponent by 1, then divide by the new exponent.
The reverse power rule
To integrate xn (where n=−1), do:
∫xndx=n+1xn+1+C
The +C is crucial. Why? Because the derivative of any constant is zero. If F(x)=31x3+5, its derivative is still x2. So when we integrate, we must add an arbitrary constant C to account for all possible original functions.
The Precise Statement
For a polynomial P(x)=anxn+an−1xn−1+⋯+a1x+a0, its indefinite integral (antiderivative) is:
∫P(x)dx=n+1anxn+1+nan−1xn+⋯+2a1x2+a0x+C
You integrate term by term, applying the reverse power rule to each term separately. The constant term a0 integrates to a0x, since ∫a0dx=a0x (because the derivative of a0x is a0).
The n=−1 exception
The reverse power rule ∫xndx=n+1xn+1 fails when n=−1, because you'd be dividing by zero. That case (∫x1dx) gives log∣x∣+C, not a power of x. For polynomials, this never arises — polynomial exponents are non-negative integers.
A Worked Example
Integrate f(x)=4x3−2x+7.
Apply the rule term by term:
- 4x3: increase exponent to 4, divide by 4 → 44x4=x4
- −2x: this is −2x1, increase exponent to 2, divide by 2 → 2−2x2=−x2
- 7: this is 7x0, increase exponent to 1, divide by 1 → 7x
So:
∫(4x3−2x+7)dx=x4−x2+7x+C
You can check by differentiating: the derivative of x4−x2+7x+C is 4x3−2x+7, which is exactly your original function.
The check
Differentiation is the proof of integration. Always verify your answer by differentiating it — you should recover the original integrand.
Definite Integration: Area Under the Curve
When you want the actual area between x=a and x=b, you use the definite integral:
∫abP(x)dx=F(b)−F(a)
where F(x) is any antiderivative of P(x). The constant C cancels out, so you can ignore it.
For example, the area under f(x)=4x3−2x+7 from x=1 to x=3:
F(x)=x4−x2+7x
F(3)=81−9+21=93
F(1)=1−1+7=7
∫13f(x)dx=93−7=86
That's the exact area — no approximations, no rectangles. Integration gives the precise value.
Why It Works (Briefly)
Integration is accumulation. The derivative measures instantaneous rate of change; the integral measures the total accumulation of that change. For a polynomial, the reverse power rule works because the derivative of n+1xn+1 is exactly xn — the two operations are inverses. That's the fundamental theorem of calculus in action: differentiation and integration undo each other.
∫xndx=n+1xn+1+C(n=−1)
∫(anxn+⋯+a0)dx=n+1anxn+1+⋯+a0x+C
Polynomial integration is the simplest entry point into calculus — a clean, mechanical process that builds directly on what you already know about derivatives. Master this, and you have the foundation for integrating everything else.
Since the integrand and its reflection under x→−x add up to the constant 1, the symmetric-interval property ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx gives the value directly.
∫−11ex+e−xexdx=1
Use the symmetric-interval property; the integrand plus its reflection sums to 1, giving I=1.
∫−aaf(x)dx=∫−aaf(−x)dx. If f(x)+f(−x)=k (constant), then 2I=∫−aakdx.
- Let I=∫−11ex+e−xexdx and f(x)=ex+e−xex.
- Replace x→−x: f(−x)=e−x+exe−x.
- Add: f(x)+f(−x)=ex+e−xex+e−x=1.
- Hence 2I=∫−11(f(x)+f(−x))dx=∫−111dx=[x]−11=2.
- ∴I=22=1.
∫−11ex+e−xexdx=1.
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