Show that ∫10sinx√x2+1dx<√2−1 ?
1 Answer
See explanation
Explanation:
We want to show
∫10sin(x)√x2+1dx<√2−1
This is a quite "ugly" integral, so our approach will not be to solve this integral, but compare it to a "nicer" integral
We now that for all positive real numbers
Thus, the value of the integrand will also be bigger, for all positive real numbers, if we substitute
∫10x√x2+1dx<√2−1
Then our first statement must also be true
The new integral is a simple substitution problem
∫10x√x2+1=[√x2+1]10=√2−1
The last step is to notice that
Therefore we can conclude
∫10sin(x)√x2+1dx<√2−1