Definite IntegrationmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Cubic with Complex Root, Definite Integral = -8 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
If α=1\alpha=1 and β=1+i2\beta=1+i\sqrt2, where i=−1i=\sqrt{-1}, are two roots of the equation x3+ax2+bx+c=0x^3+ax^2+bx+c=0, a,b,c∈Ra,b,c\in\mathbb{R}, then ∫−11(x3+ax2+bx+c)dx\displaystyle\int_{-1}^{1}\left(x^3+ax^2+bx+c\right)dx is equal to
A−2-2
B−4-4
C−8-8correct
D−10-10
Solution
Step 1: Since the coefficients are real, the complex root 1+i21+i\sqrt2 comes with its conjugate 1−i21-i\sqrt2. So the three roots are 1, 1+i2, 1−i21,\ 1+i\sqrt2,\ 1-i\sqrt2. Step 2: Sum of roots =−a=-a:
1+(1+i2)+(1−i2)=3 ⇒ a=−3.1+(1+i\sqrt2)+(1-i\sqrt2)=3\ \Rightarrow\ a=-3.
 Step 3: Product of roots =−c=-c:
1⋅(1+i2)(1−i2)=1⋅(1+2)=3 ⇒ c=−3.1\cdot(1+i\sqrt2)(1-i\sqrt2)=1\cdot(1+2)=3\ \Rightarrow\ c=-3.
 Step 4: The integral (the x3x^3 and bxbx terms are odd, so vanish over [−1,1][-1,1]):
I=∫−11(x3−3x2+bx−3) dx=2∫01(−3x2−3) dx=2[−x3−3x]01=2(−1−3)=−8.I=\int_{-1}^{1}(x^3-3x^2+bx-3)\,dx=2\int_0^1(-3x^2-3)\,dx=2\big[-x^3-3x\big]_0^1=2(-1-3)=-8.
 Correct answer: (3)
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