Q.Evaluate the following:
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.
Each integral uses the linear-substitution rule ∫f(px+q)dx=p1F(px+q) for a power, exponential, or (in part (d), after expanding the square) exponential term.
(a) 31(2x+3)3/2+C (b) −51e4−5x+C (c) 3a(ax+b)3+C (d) 2logaa2x−a−2x+2x+C.
Each uses the linear-substitution rule ∫f(px+q)dx=p1F(px+q); part (d) is expanded first.
∫(px+q)ndx=p(n+1)(px+q)n+1+C,∫epx+qdx=pepx+q+C,∫akxdx=klogaakx+C.
Steps
- (a) p=2, n=21:
∫(2x+3)1/2dx=2⋅23(2x+3)3/2=31(2x+3)3/2+C.
- (b) p=−5:
∫e4−5xdx=−5e4−5x=−51e4−5x+C.
- (c) p=a, n=2:
∫(ax+b)2dx=a⋅3(ax+b)3=3a(ax+b)3+C.
- (d) Expand: (ax+a−x)2=a2x+2+a−2x.
∫(a2x+2+a−2x)dx=2logaa2x+2x+−2logaa−2x=2logaa2x−a−2x+2x+C.
- 31(2x+3)3/2+C;
- −51e4−5x+C;
- 3a(ax+b)3+C;
- 2logaa2x−a−2x+2x+C.
- CBSE 2025Set 465/W1XZY/41 markMCQQ.Assertion (A) : The area of the region bounded by the line y−1=x, the x-axis and the ordinates x=−1 and x=1 is 2 square units. Reason (R) : The area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b is given by ∫abf(x)dx. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution
∫−11(x+1)dx=2, so the Assertion is true; the Reason gives exactly the formula used, so it correctly explains it.
Area =∫abf(x)dx for y=f(x)≥0 between x=a and x=b.
- Rewrite the line: y−1=x⇒y=x+1; over [−1,1] it is ≥0 (it touches the axis at x=−1).
- Set up the area: Area=∫−11(x+1)dx.
- Integrate: [2x2+x]−11=(21+1)−(21−1)=23−(−21)=2.
- So the Assertion (area =2) is true, and the Reason is the exact general formula applied here, making it the correct explanation.
✓Final answerBoth Assertion and Reason are true, and the Reason correctly explains the Assertion — option (A).
- CBSE 2024Set 465/S/RQPS/41 markMCQQ.∫22x⋅3xdx is equal to : (A) log1212x+C (B) log2⋅log322x⋅3x+C (C) log64⋅6x+C (D) (12)x⋅log12+C
›Reveal solutionSolution
22x3x=12x, and ∫axdx=logaax+C, giving log1212x+C.
∫axdx=logaax+C, where a>0, a=1.
- Simplify the integrand: 22x=(22)x=4x, so 22x⋅3x=4x⋅3x=(4⋅3)x=12x.
- Integrate: ∫12xdx=log1212x+C.
✓Final answer(A) log1212x+C
- CBSE 2023Set 465/EF1GH/41 markMCQQ.∫(x−1)e−xdx is equal to :(a) (x−2)e−x+C(b) xe−x+C(c) −xe−x+C(d) (x+1)e−x+C
›Reveal solutionSolution
∫(x−1)e−xdx=−xe−x+C, verified by differentiation.
Integration by parts: ∫udv=uv−∫vdu. Here take u=(x−1), dv=e−xdx.
- With u=x−1⇒du=dx and dv=e−xdx⇒v=−e−x.
- ∫(x−1)e−xdx=(x−1)(−e−x)−∫(−e−x)dx=−(x−1)e−x−e−x+C.
- Simplify: =−(x−1)e−x−e−x=−e−x[(x−1)+1]=−xe−x+C.
- Check by differentiating: dxd(−xe−x)=−e−x+xe−x=(x−1)e−x ✓.
✓Final answer(c) −xe−x+C
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