Q.Evaluate the definite integral: ∫12(4x3−5x2+6x+9) dx
Concept understanding — Power Rule Integration
The Power Rule for Integration
Integration reverses differentiation: given a rate of change, it recovers the original function. When you differentiate xn you get nxn−1 — the exponent drops by one and multiplies in front. To integrate you do the opposite: raise the exponent by one and divide by the new exponent. That is the whole idea.
The statement
∫xndx=n+1xn+1+C,n=−1
- n may be any real number except −1 (fractions, negatives and 0 all work).
- C is the constant of integration — shifting a graph up or down does not change its slope, so infinitely many functions share the same derivative.
Why it works
Differentiate the answer and you should get back the integrand:
dxd(n+1xn+1+C)=n+1(n+1)xn=xn.
That one line is the proof.
Using it
∫x3dx=4x4+C,∫xdx=∫x1/2dx=3/2x3/2+C=32x3/2+C.
For a polynomial, apply it term by term:
∫(5x3−2x+7)dx=45x4−x2+7x+C.
The one exception: n=−1
The formula needs n+1=0. For n=−1 it would divide by zero, so a different result takes over:
∫x1dx=log∣x∣+C.
The absolute value keeps the logarithm defined for negative x as well.
The commonest slip is forgetting to divide by the new exponent — writing ∫x3dx=x4+C. Check by differentiating: dxdx4=4x3, not x3, so you must divide by 4.
The power rule for integration is the very first formula taught in the NCERT Class 12 Integrals chapter and underlies nearly every subsequent integration technique in CBSE boards and JEE Main. Students searching 'power rule of integration class 12 formula' or 'integration of xn examples' will find this raise-the-exponent-and-divide method, along with its log|x| exception at n = -1, is exactly what board exams test first.
Concept: Definite Integral — Power Rule & Linear Combination
We integrate term-by-term using ∫xndx=n+1xn+1 and then evaluate from 1 to 2.
Step 1: Find the antiderivative
∫(4x3−5x2+6x+9)dx=4⋅4x4−5⋅3x3+6⋅2x2+9x=x4−35x3+3x2+9x.
Step 2: Evaluate at the limits
At x=2:
24−35(8)+3(4)+18=16−340+12+18=46−340=3138−40=398.
At x=1:
1−35+3+9=13−35=339−5=334.
Step 3: Subtract
398−334=364.
The value is 364.
Apply the power rule term by term and evaluate between the limits. The value is 364.
Step-by-step solution
1. Find the antiderivative.
F(x)=∫(4x3−5x2+6x+9)dx=x4−35x3+3x2+9x.
2. Evaluate at the upper limit x=2.
F(2)=16−35(8)+3(4)+18=46−340=398.
3. Evaluate at the lower limit x=1.
F(1)=1−35(1)+3(1)+9=13−35=334.
4. Subtract.
∫12(4x3−5x2+6x+9)dx=F(2)−F(1)=398−334=364.
∫12(4x3−5x2+6x+9)dx=364
Method: Definite integral of a polynomial (term-by-term power rule)
Integrate each power of x separately, then evaluate F(b)−F(a).
Steps
Step 1: Apply ∫xndx=n+1xn+1 to every term.
For 4x3−5x2+6x+9: F(x)=x4−35x3+3x2+9x.
Step 2: Form the evaluation bracket [F(x)]ab.
Step 3: Substitute the upper and lower limits and subtract, keeping fractions exact.
Step 4: Combine to a single value. No +C for a definite integral; a common denominator tidies the fractions.
Common Mistakes
Mistake 1: Antiderivative of −5x2 taken as −5x3 (forgetting ÷3).
Why it's wrong: ∫−5x2dx=−35x3. Correct approach: divide by the new exponent.
Mistake 2: Antiderivative of the constant 9 dropped.
Why it's wrong: ∫9dx=9x, a real contribution. Correct approach: integrate constants to 9x.
Mistake 3: Arithmetic slip in F(2)−F(1) with mixed fractions.
Why it's wrong: careless subtraction gives a wrong number. Correct approach: convert to a common denominator, e.g. 398−334=364.
- KEAM 2026Set eng-2026-04174 marksMCQQ.If ∫x5x(4x2−1)dx=8x5+ax+C, where a is a constant and C is the constant of integration, then the value of a is (A) 20 (B) −20 (C) −10 (D) 10 (E) −8
›Reveal solutionSolution
Simplify the integrand to powers of x and integrate.
∫x5x(4x2−1)dx=5∫x−1/2(4x2−1)dx=5∫(4x3/2−x−1/2)dx.
Integrating:
=5[4⋅52x5/2−2x1/2]=8x5/2−10x1/2+C=8x5−10x+C.
Comparing with 8x5+ax+C gives a=−10.
✓Final answerThe correct option is (C).
- KEAM 2026Set eng-2026-04214 marksMCQQ.∫sin3xelogcosxdx= (A) 4cos4x+C (B) 4−cos4x+C (C) 4xcos4x+C (D) 4sin4x+C (E) 4−sin4x+C
›Reveal solutionSolution
elogcosx=cosx, then substitute u=sinx to get 4sin4x+C.
elogcosx=cosx, so the integrand is sin3xcosx.
Let u=sinx, du=cosxdx: ∫u3du=4u4+C=4sin4x+C.
✓Final answerThe correct option is (D).
- KEAM 2025Set eng-2025-04294 marksMCQQ.∫x+x−2dx is equal to (A) 21(x3/2−(x−1)3/2)+C (B) 31(x3/2−(x−2)3/2)+C (C) 31(x2/3−(1−x)2/3)+C (D) 21(x2/3−(1−x)2/3)+C (E) 31(x2/3−(x−2)2/3)+C
›Reveal solutionSolution
Rationalizing the denominator (difference =2) gives 21(x−x−2); integrating term by term yields 31(x3/2−(x−2)3/2)+C.
Multiply numerator and denominator by x−x−2:
x+x−21=x−(x−2)x−x−2=2x−x−2.
Integrate:
21∫(x1/2−(x−2)1/2)dx=21⋅32(x3/2−(x−2)3/2)+C=31(x3/2−(x−2)3/2)+C.
✓Final answerThe correct option is (B).
- KEAM 2025Set eng-2025-04294 marksMCQQ.If f′(x)=3x2−x32 and f(1)=0, then f(x)= (A) x2+x31+1 (B) x3+x21+1 (C) x3+x21+2 (D) x3+x21−2 (E) x3+x21−1
›Reveal solutionSolution
Antidifferentiating 3x2−2x−3 gives x3+x−2+C; using f(1)=0 fixes C=−2.
Given f′(x)=3x2−x32=3x2−2x−3, integrate:
f(x)=x3−2⋅−2x−2+C=x3+x21+C.
Apply f(1)=0:
1+1+C=0⇒C=−2.
Therefore f(x)=x3+x21−2.
✓Final answerThe correct option is (D).
- KEAM 2024Set eng-2024-06094 marksMCQQ.∫e4logx−e3logxe6logx−e5logxdx= (A) ex+C (B) 2x2+C (C) x+C (D) 3x3+C (E) xex+C
›Reveal solutionSolution
Convert eklogx=xk; the fraction reduces to x2, whose integral is 3x3+C.
Since eklogx=xk,
e4logx−e3logxe6logx−e5logx=x4−x3x6−x5=x3(x−1)x5(x−1)=x2.
Hence ∫x2dx=3x3+C.
✓Final answerThe correct option is (D).
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