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Question: If $I_1 = \int_0^1 (1-(1-x^3)^{\sqrt{2}})^{\sqrt{3}} x^2 dx$ and $I_2 = \int_0^1 (1-(1-x^3)^{\sqrt{2...

If I1=∫01(1−(1−x3)2)3x2dxI_1 = \int_0^1 (1-(1-x^3)^{\sqrt{2}})^{\sqrt{3}} x^2 dx and I2=∫01(1−(1−x3)2)3+1x2dxI_2 = \int_0^1 (1-(1-x^3)^{\sqrt{2}})^{\sqrt{3}+1} x^2 dx, then I1I2−3−122+0.210\frac{\frac{I_1}{I_2} - \frac{\sqrt{3}-1}{2\sqrt{2}} + 0.2}{10} is equal to

Answer

0.12

Explanation

Solution

Solution:

We are given

I1=∫01(1−(1−x3)2)3x2 dx,I2=∫01(1−(1−x3)2)3+1x2 dx.I_1=\int_{0}^{1}\Bigl(1-(1-x^3)^{\sqrt{2}}\Bigr)^{\sqrt{3}} x^2\,dx,\quad I_2=\int_{0}^{1}\Bigl(1-(1-x^3)^{\sqrt{2}}\Bigr)^{\sqrt{3}+1} x^2\,dx.

Step 1. Change of Variable

Let t=x3t=x^3. Then,

dt=3x2dx⟹x2dx=dt3,dt=3x^2dx\quad\Longrightarrow\quad x^2dx=\frac{dt}{3},

and when x=0, t=0x=0,\,t=0 and x=1, t=1x=1,\,t=1. So,

I1=13∫01(1−(1−t)2)3 dt,I2=13∫01(1−(1−t)2)3+1 dt.I_1=\frac{1}{3}\int_{0}^{1}\Bigl(1-(1-t)^{\sqrt{2}}\Bigr)^{\sqrt{3}}\,dt,\quad I_2=\frac{1}{3}\int_{0}^{1}\Bigl(1-(1-t)^{\sqrt{2}}\Bigr)^{\sqrt{3}+1}\,dt.

Now let u=1−tu=1-t so that dt=−dudt=-du and the limits change: t=0⇒u=1t=0\Rightarrow u=1, t=1⇒u=0t=1\Rightarrow u=0. Reversing the limits:

I1=13∫01(1−u2)3 du,I2=13∫01(1−u2)3+1 du.I_1=\frac{1}{3}\int_{0}^{1}(1-u^{\sqrt{2}})^{\sqrt{3}}\,du,\quad I_2=\frac{1}{3}\int_{0}^{1}(1-u^{\sqrt{2}})^{\sqrt{3}+1}\,du.

Step 2. Secondary Substitution and Beta Integral

Now set v=u2v=u^{\sqrt{2}} so that

u=v12anddu=12 v12−1dv.u=v^{\frac{1}{\sqrt{2}}}\quad\text{and}\quad du=\frac{1}{\sqrt{2}}\,v^{\frac{1}{\sqrt{2}}-1}dv.

Then,

I1=132∫01(1−v)3v12−1 dv,I_1=\frac{1}{3\sqrt{2}}\int_{0}^{1}(1-v)^{\sqrt{3}} v^{\frac{1}{\sqrt{2}}-1}\,dv, I2=132∫01(1−v)3+1v12−1 dv.I_2=\frac{1}{3\sqrt{2}}\int_{0}^{1}(1-v)^{\sqrt{3}+1} v^{\frac{1}{\sqrt{2}}-1}\,dv.

Recognize these as Beta integrals:

∫01vp−1(1−v)q−1dv=B(p,q)=Γ(p)Γ(q)Γ(p+q).\int_{0}^{1}v^{p-1}(1-v)^{q-1}dv = B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.

Here for I1I_1: p=12, q=3+1p=\frac{1}{\sqrt{2}},\ q=\sqrt{3}+1; for I2I_2: p=12, q=3+2p=\frac{1}{\sqrt{2}},\ q=\sqrt{3}+2.

Thus,

I1=132Γ(12)Γ(3+1)Γ(12+3+1),I_1=\frac{1}{3\sqrt{2}}\frac{\Gamma\left(\tfrac{1}{\sqrt{2}}\right)\Gamma\left(\sqrt{3}+1\right)} {\Gamma\left(\frac{1}{\sqrt{2}}+\sqrt{3}+1\right)}, I2=132Γ(12)Γ(3+2)Γ(12+3+2).I_2=\frac{1}{3\sqrt{2}}\frac{\Gamma\left(\tfrac{1}{\sqrt{2}}\right)\Gamma\left(\sqrt{3}+2\right)} {\Gamma\left(\frac{1}{\sqrt{2}}+\sqrt{3}+2\right)}.

Step 3. Compute the Ratio I1I2 \dfrac{I_1}{I_2}

I1I2=B(12,3+1)B(12,3+2)=Γ(3+1)Γ(3+2)⋅Γ(12+3+2)Γ(12+3+1).\frac{I_1}{I_2}=\frac{B\left(\frac{1}{\sqrt{2}},\sqrt{3}+1\right)}{B\left(\frac{1}{\sqrt{2}},\sqrt{3}+2\right)} =\frac{\Gamma(\sqrt{3}+1)}{\Gamma(\sqrt{3}+2)} \cdot\frac{\Gamma\left(\frac{1}{\sqrt{2}}+\sqrt{3}+2\right)} {\Gamma\left(\frac{1}{\sqrt{2}}+\sqrt{3}+1\right)}.

Using the property Γ(z+1)=z Γ(z)\Gamma(z+1)=z\,\Gamma(z):

Γ(3+2)=(3+1) Γ(3+1),\Gamma(\sqrt{3}+2)=(\sqrt{3}+1)\,\Gamma(\sqrt{3}+1), Γ(12+3+2)=(12+3+1)Γ(12+3+1).\Gamma\left(\frac{1}{\sqrt{2}}+\sqrt{3}+2\right)=\Bigl(\frac{1}{\sqrt{2}}+\sqrt{3}+1\Bigr) \Gamma\left(\frac{1}{\sqrt{2}}+\sqrt{3}+1\right).

Thus,

I1I2=13+1(12+3+1)=1+12(3+1).\frac{I_1}{I_2}=\frac{1}{\sqrt{3}+1}\Bigl(\frac{1}{\sqrt{2}}+\sqrt{3}+1\Bigr) =1+\frac{1}{\sqrt{2}(\sqrt{3}+1)}.

Step 4. Evaluate the Final Expression

We need to compute

E=I1I2−3−122+0.210.E=\frac{\frac{I_1}{I_2}-\frac{\sqrt{3}-1}{2\sqrt{2}}+0.2}{10}.

Substitute I1I2=1+12(3+1)\dfrac{I_1}{I_2}=1+\frac{1}{\sqrt{2}(\sqrt{3}+1)}:

E=1+12(3+1)−3−122+0.210.E=\frac{1+\frac{1}{\sqrt{2}(\sqrt{3}+1)}-\frac{\sqrt{3}-1}{2\sqrt{2}}+0.2}{10}.

Notice that

12(3+1)=222(3+1)\frac{1}{\sqrt{2}(\sqrt{3}+1)}=\frac{2}{2\sqrt{2}(\sqrt{3}+1)}

and

3−122.\frac{\sqrt{3}-1}{2\sqrt{2}}.

Observe that:

222(3+1)−3−122=2−(3−1)(3+1)22(3+1).\frac{2}{2\sqrt{2}(\sqrt{3}+1)} - \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{2 - (\sqrt{3}-1)(\sqrt{3}+1)}{2\sqrt{2}(\sqrt{3}+1)}.

But (3−1)(3+1)=3−1=2(\sqrt{3}-1)(\sqrt{3}+1)=3-1=2. Thus the fraction is

2−222(3+1)=0.\frac{2-2}{2\sqrt{2}(\sqrt{3}+1)}=0.

So the expression reduces to:

E=1+0.210=1.210=0.12.E=\frac{1+0.2}{10}=\frac{1.2}{10}=0.12.

Summary:

  • Explanation: Change of variables transforms the integrals to Beta functions. Taking the ratio and using the Gamma function property simplifies the ratio to 1+12(3+1)1+\frac{1}{\sqrt{2}(\sqrt{3}+1)}; the additional terms cancel out, yielding the final answer 0.12.