This is an MCQ question where one needs to solve it, without a calculator, within 3 minutes. I know Simpson's rule, trapezoidal rule, etc. But it may need a different clever way without actually evaluating numerically.
#Numerical estimation for an integral, without a calculator
placid gulch
gritty epochBOT
placid gulch
<@&286206848099549185>
prime fog
the integrand is (very) approximately 1+x^2, so I(a,b) = (b + (b^3)/3) - (a + (a^3)/3) is a reasonable approximation
uncut heraldBOT
Unique
I would agree with thar only when $x \le 1$.
Can you explain more why would you neglect other terms since we have values of x higher than 1?
honest holly
It's a different kind of approximation
x^3 + x is not too far from x^2 when x^4 is around
placid gulch
x^3 + x is not too far from x^2 when x^4 is around
I still did not get it
placid gulch
<@&286206848099549185>
honest holly
Try plugging in x= 2, then 3, then 4, then 5 into the integrand and see what the numbers look like
placid gulch
Try plugging in x= 2, then 3, then 4, then 5 into the integrand and see what the...
Plugging $x=2$ is easy, then for other values, it will take a long time. We need an approach to solve it in a short time
uncut heraldBOT
Unique
honest holly
the integrand is (very) approximately 1+x^2, so I(a,b) = (b + (b^3)/3) - (a + (a^3)/3) is a reasonable approximation
What was wrong with this?
placid gulch
the integrand is (very) approximately 1+x^2, so I(a,b) = (b + (b^3)/3) - (a + (a...
That I do not know how we got that approximation
honest holly
1 + x + x^2 + x^3 + x^4 is pretty close to 1 + 2x^2 + x^4