The value of \int\limits_0^1(2 x^3-3 x^2-x+1)^\frac13 ~d x is equal to :

Definite Integration PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q46 with Solution

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Correct answer: NONE OF THE GIVEN OPTIONS IS CORRECT; THE INTEGRAL IS APPROXIMATELY 0.71, NOT 0.

Let $I=\int_0^1 \left(2x^3-3x^2-x+1\right)^{1/3}\,dx.$
First factor the cubic inside the cube root: $2x^3-3x^2-x+1=(x-1)(2x^2-x-1).$ Now, $2x^2-x-1=(2x+1)(x-1).$ Hence, $2x^3-3x^2-x+1=(x-1)^2(2x+1).$ So the integral becomes

=\int_0^1 (1-x)^{2/3}(2x+1)^{1/3}\,dx.$$ 3. Now use the substitution $$t=1-x \quad\Rightarrow\quad x=1-t,\quad dx=-dt.$$ When $x=0$, $t=1$; when $x=1$, $t=0$. Thus $$I=\int_1^0 t^{2/3}\big(2(1-t)+1\big)^{1/3}(-dt) =\int_0^1 t^{2/3}(3-2t)^{1/3}\,dt.$$ This is not yet immediately one of the options, so we look for a better substitution. 4. Instead, set $$u=\frac{1-x}{(2x+1)^{1/2}}$$ which is not convenient. A more direct approach is to notice a special algebraic form. Let us try the substitution $$1-x=t^3 \Rightarrow x=1-t^3,\quad dx=-3t^2dt.$$ Then $$(1-x)^{2/3}=t^2, \qquad 2x+1=2(1-t^3)+1=3-2t^3.$$ So $$I=\int_1^0 t^2(3-2t^3)^{1/3}(-3t^2dt) =3\int_0^1 t^4(3-2t^3)^{1/3}dt,$$ which again is not the simplest route. 5. Let us instead search for an antiderivative by recognizing $$2x^3-3x^2-x+1=(1-x)^2(2x+1).$$ Consider $$y=(1-x)^{2/3}(2x+1)^{1/3}.$$ We test whether it can be written in a form suitable for substitution with $$u=\frac{1-x}{2x+1}.$$ Then $$u=\frac{1-x}{2x+1}, \qquad 1-x=u(2x+1).$$ Differentiate: $$du=\frac{-(2x+1)-2(1-x)}{(2x+1)^2}dx=\frac{-3}{(2x+1)^2}dx,

so $dx=-\frac{(2x+1)^2}{3}du.$ Also, $1-x=u(2x+1).$ Thus $(1-x)^{2/3}(2x+1)^{1/3}=u^{2/3}(2x+1).$ Hence $I=\int u^{2/3}(2x+1)\left(-\frac{(2x+1)^2}{3}du\right),$ which still needs $x$ in terms of $u$ . From $u=\frac{1-x}{2x+1} \Rightarrow x=\frac{1-u}{1+2u},$ so $2x+1=\frac{3}{1+2u}.$ Therefore

=-9\int \frac{u^{2/3}}{(1+2u)^3}du.$$ The limits are: when $x=0$, $u=1$; when $x=1$, $u=0$. So $$I=9\int_0^1 \frac{u^{2/3}}{(1+2u)^3}du.$$ This is positive, so options $-1$ and $0$ are unlikely. 6. Let us evaluate numerically to identify the correct option. Since $$f(x)=\big((1-x)^2(2x+1)\big)^{1/3} \ge 0 \quad \text{for } x\in[0,1],$$ we must have $$I\ge 0.$$ Also, $$f(0)=1, \qquad f(1)=0.$$ Take a few values: $$f\left(\tfrac12\right)=\left(2\cdot \tfrac18-3\cdot \tfrac14-\tfrac12+1\right)^{1/3}=0^{1/3}=0,$$ actually checking carefully, $$2\cdot\frac18-3\cdot\frac14-\frac12+1=\frac14-\frac34-\frac12+1=0,$$ so yes, $f(1/2)=0$. Now use factorization: $$f(x)=(1-x)^{2/3}(2x+1)^{1/3}.$$ This is positive except at $x=1$. So the integral is definitely not $0$. 7. We now test option $D=1$ by finding the exact antiderivative. Observe that $$\frac{d}{dx}\left[\frac{1}{2}(1-x)^{2/3}(2x+1)^{4/3}\right] =\frac12\left[-\frac23(1-x)^{-1/3}(2x+1)^{4/3}+\frac83(1-x)^{2/3}(2x+1)^{1/3}\right].$$ Factoring, $$=\frac13(1-x)^{-1/3}(2x+1)^{1/3}\left[-(2x+1)+4(1-x)\right] =\frac13(1-x)^{-1/3}(2x+1)^{1/3}(3-6x).$$ This does not match directly. Try instead $$F(x)=-(1-x)^{2/3}(2x+1)^{1/3}.$$ Then $$F'(x)=\frac23(1-x)^{-1/3}(2x+1)^{1/3}-\frac23(1-x)^{2/3}(2x+1)^{-2/3},$$ not equal either. 8. Since the integrand is positive and bounded by $1$ on most of the interval, the integral should be between $0$ and $1$. Among the given options, only $1$ is plausible as a positive value. To check, use Simpson’s rule on $$f(x)=\big((1-x)^2(2x+1)\big)^{1/3}.$$ At points $x=0,\frac14,\frac12,\frac34,1$: $$f(0)=1,$$ $$f\left(\frac14\right)=\left(\frac{9}{16}\cdot\frac32\right)^{1/3}=\left(\frac{27}{32}\right)^{1/3}\approx 0.9449,$$ $$f\left(\frac12\right)=\left(\frac14\cdot2\right)^{1/3}=\left(\frac12\right)^{1/3}\approx 0.7937,$$ using the factorized form correctly, $$(1-1/2)^2(2\cdot1/2+1)=\frac14\cdot2=\frac12,$$ so earlier direct arithmetic was mistaken. Similarly, $$f\left(\frac34\right)=\left(\frac1{16}\cdot\frac52\right)^{1/3}=\left(\frac5{32}\right)^{1/3}\approx 0.5386,$$ $$f(1)=0.$$ Then Simpson’s estimate: $$I\approx \frac{1/4}{3}\left[f(0)+f(1)+4\left(f\left(\tfrac14\right)+f\left(\tfrac34\right)\right)+2f\left(\tfrac12\right)\right].$$ $$I\approx \frac1{12}\left[1+0+4(0.9449+0.5386)+2(0.7937)\right] =\frac1{12}(1+5.934+1.5874) \approx \frac{8.5214}{12} \approx 0.71.$$ So the integral is not exactly any of $0,1,2,-1$, but among options it is certainly not $0$. 9. Let us compute exactly by substituting $$x=\frac{1-u}{2},\quad dx=-\frac12du.$$ Then $$1-x=\frac{1+u}{2},\qquad 2x+1=2-u+1=2-u?$$ This is not simplifying well. A much better substitution is $$2x+1=3t \Rightarrow x=\frac{3t-1}{2},\quad dx=\frac32dt.$$ Then $$1-x=1-\frac{3t-1}{2}=\frac{3(1-t)}{2}.$$ Hence $$f(x)=(1-x)^{2/3}(2x+1)^{1/3}=\left(\frac{3(1-t)}{2}\right)^{2/3}(3t)^{1/3} =\frac{3}{2^{2/3}}(1-t)^{2/3}t^{1/3}.$$ Also $x:0\to1$ gives $t:\frac13\to1$. So $$I=\frac32\cdot \frac{3}{2^{2/3}}\int_{1/3}^1 (1-t)^{2/3}t^{1/3}dt,$$ again positive and clearly nonzero. Thus the stored answer $C:0$ is incorrect. 10. Since none of the options matches the actual value (approximately $0.71$), the question/options seem flawed. The integral is certainly **not** $0$. Therefore, I disagree with the stored answer.

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Correct answer: NONE OF THE GIVEN OPTIONS IS CORRECT; THE INTEGRAL IS APPROXIMATELY 0.71, NOT 0.

Let $I=\int_0^1 \left(2x^3-3x^2-x+1\right)^{1/3}\,dx.$
First factor the cubic inside the cube root: $2x^3-3x^2-x+1=(x-1)(2x^2-x-1).$ Now, $2x^2-x-1=(2x+1)(x-1).$ Hence, $2x^3-3x^2-x+1=(x-1)^2(2x+1).$ So the integral becomes

=\int_0^1 (1-x)^{2/3}(2x+1)^{1/3}\,dx.$$ 3. Now use the substitution $$t=1-x \quad\Rightarrow\quad x=1-t,\quad dx=-dt.$$ When $x=0$, $t=1$; when $x=1$, $t=0$. Thus $$I=\int_1^0 t^{2/3}\big(2(1-t)+1\big)^{1/3}(-dt) =\int_0^1 t^{2/3}(3-2t)^{1/3}\,dt.$$ This is not yet immediately one of the options, so we look for a better substitution. 4. Instead, set $$u=\frac{1-x}{(2x+1)^{1/2}}$$ which is not convenient. A more direct approach is to notice a special algebraic form. Let us try the substitution $$1-x=t^3 \Rightarrow x=1-t^3,\quad dx=-3t^2dt.$$ Then $$(1-x)^{2/3}=t^2, \qquad 2x+1=2(1-t^3)+1=3-2t^3.$$ So $$I=\int_1^0 t^2(3-2t^3)^{1/3}(-3t^2dt) =3\int_0^1 t^4(3-2t^3)^{1/3}dt,$$ which again is not the simplest route. 5. Let us instead search for an antiderivative by recognizing $$2x^3-3x^2-x+1=(1-x)^2(2x+1).$$ Consider $$y=(1-x)^{2/3}(2x+1)^{1/3}.$$ We test whether it can be written in a form suitable for substitution with $$u=\frac{1-x}{2x+1}.$$ Then $$u=\frac{1-x}{2x+1}, \qquad 1-x=u(2x+1).$$ Differentiate: $$du=\frac{-(2x+1)-2(1-x)}{(2x+1)^2}dx=\frac{-3}{(2x+1)^2}dx,

=-9\int \frac{u^{2/3}}{(1+2u)^3}du.$$ The limits are: when $x=0$, $u=1$; when $x=1$, $u=0$. So $$I=9\int_0^1 \frac{u^{2/3}}{(1+2u)^3}du.$$ This is positive, so options $-1$ and $0$ are unlikely. 6. Let us evaluate numerically to identify the correct option. Since $$f(x)=\big((1-x)^2(2x+1)\big)^{1/3} \ge 0 \quad \text{for } x\in[0,1],$$ we must have $$I\ge 0.$$ Also, $$f(0)=1, \qquad f(1)=0.$$ Take a few values: $$f\left(\tfrac12\right)=\left(2\cdot \tfrac18-3\cdot \tfrac14-\tfrac12+1\right)^{1/3}=0^{1/3}=0,$$ actually checking carefully, $$2\cdot\frac18-3\cdot\frac14-\frac12+1=\frac14-\frac34-\frac12+1=0,$$ so yes, $f(1/2)=0$. Now use factorization: $$f(x)=(1-x)^{2/3}(2x+1)^{1/3}.$$ This is positive except at $x=1$. So the integral is definitely not $0$. 7. We now test option $D=1$ by finding the exact antiderivative. Observe that $$\frac{d}{dx}\left[\frac{1}{2}(1-x)^{2/3}(2x+1)^{4/3}\right] =\frac12\left[-\frac23(1-x)^{-1/3}(2x+1)^{4/3}+\frac83(1-x)^{2/3}(2x+1)^{1/3}\right].$$ Factoring, $$=\frac13(1-x)^{-1/3}(2x+1)^{1/3}\left[-(2x+1)+4(1-x)\right] =\frac13(1-x)^{-1/3}(2x+1)^{1/3}(3-6x).$$ This does not match directly. Try instead $$F(x)=-(1-x)^{2/3}(2x+1)^{1/3}.$$ Then $$F'(x)=\frac23(1-x)^{-1/3}(2x+1)^{1/3}-\frac23(1-x)^{2/3}(2x+1)^{-2/3},$$ not equal either. 8. Since the integrand is positive and bounded by $1$ on most of the interval, the integral should be between $0$ and $1$. Among the given options, only $1$ is plausible as a positive value. To check, use Simpson’s rule on $$f(x)=\big((1-x)^2(2x+1)\big)^{1/3}.$$ At points $x=0,\frac14,\frac12,\frac34,1$: $$f(0)=1,$$ $$f\left(\frac14\right)=\left(\frac{9}{16}\cdot\frac32\right)^{1/3}=\left(\frac{27}{32}\right)^{1/3}\approx 0.9449,$$ $$f\left(\frac12\right)=\left(\frac14\cdot2\right)^{1/3}=\left(\frac12\right)^{1/3}\approx 0.7937,$$ using the factorized form correctly, $$(1-1/2)^2(2\cdot1/2+1)=\frac14\cdot2=\frac12,$$ so earlier direct arithmetic was mistaken. Similarly, $$f\left(\frac34\right)=\left(\frac1{16}\cdot\frac52\right)^{1/3}=\left(\frac5{32}\right)^{1/3}\approx 0.5386,$$ $$f(1)=0.$$ Then Simpson’s estimate: $$I\approx \frac{1/4}{3}\left[f(0)+f(1)+4\left(f\left(\tfrac14\right)+f\left(\tfrac34\right)\right)+2f\left(\tfrac12\right)\right].$$ $$I\approx \frac1{12}\left[1+0+4(0.9449+0.5386)+2(0.7937)\right] =\frac1{12}(1+5.934+1.5874) \approx \frac{8.5214}{12} \approx 0.71.$$ So the integral is not exactly any of $0,1,2,-1$, but among options it is certainly not $0$. 9. Let us compute exactly by substituting $$x=\frac{1-u}{2},\quad dx=-\frac12du.$$ Then $$1-x=\frac{1+u}{2},\qquad 2x+1=2-u+1=2-u?$$ This is not simplifying well. A much better substitution is $$2x+1=3t \Rightarrow x=\frac{3t-1}{2},\quad dx=\frac32dt.$$ Then $$1-x=1-\frac{3t-1}{2}=\frac{3(1-t)}{2}.$$ Hence $$f(x)=(1-x)^{2/3}(2x+1)^{1/3}=\left(\frac{3(1-t)}{2}\right)^{2/3}(3t)^{1/3} =\frac{3}{2^{2/3}}(1-t)^{2/3}t^{1/3}.$$ Also $x:0\to1$ gives $t:\frac13\to1$. So $$I=\frac32\cdot \frac{3}{2^{2/3}}\int_{1/3}^1 (1-t)^{2/3}t^{1/3}dt,$$ again positive and clearly nonzero. Thus the stored answer $C:0$ is incorrect. 10. Since none of the options matches the actual value (approximately $0.71$), the question/options seem flawed. The integral is certainly **not** $0$. Therefore, I disagree with the stored answer.

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