1. Given function

We need to evaluate

and then find $4I-1$.

2. Evaluate $f(x^2)$ on $[-1,2]$

Since $x\in[-1,2]$, we have $x^2\in[0,4]$.

• If $x^2\le 2$, then $f(x^2)=[x^2]$.
• If $x^2>2$, then $f(x^2)=0$.

Now:

• For $0\le x^2<1$, $[x^2]=0$.
• For $1\le x^2\le 2$, $[x^2]=1$.
• For $x^2>2$, $f(x^2)=0$.

Thus $f(x^2)=1$ exactly when

and $f(x^2)=0$ elsewhere.

So the integrand is nonzero only on

3. Evaluate $f(x+1)$

For $x\in[-1,2]$, we have $x+1\in[0,3]$.

Now

• if $x+1\le 2$, i.e. $x\le 1$, then $f(x+1)=[x+1];$
• if $x+1>2$, i.e. $x>1$, then $f(x+1)=0.$

We only need this on the intervals where $f(x^2)=1$.

(i) On $[-\sqrt2,-1]$

Here $x\le -1$, so $x+1\in[-0.414...,0]$. Hence

(the single point $x=-1$ does not affect the integral). Therefore

So on $[-\sqrt2,-1]$, the integrand becomes simply $x$.

(ii) On $[1,\sqrt2]$

Here $x>1$, so $x+1>2$, hence

and therefore

So on $[1,\sqrt2]$, the integrand becomes

4. Split the integral

Thus,

Now compute each part.

First integral

Second integral

Therefore,

5. Find $4I-1$

6. Comparison with stored answer

Stored correct answer is $0$, but our derived value is

So the stored answer appears to be incorrect.