2:10 Wrong. There is nothing in the question to suggest that x must be positive, so √(x²) = |x| , not x. Looking at the graph of the integrand, it is positive everywhere it's defined. So any interval where it's defined will have a positive definite integral. But if we try to use the expression boxed at 8:14 to calculate the area under the curve from (for example) -20 to -10 we get a (wrongly) negative answer. This video is a good example of what can go wrong when not thinking clearly about roots of powers! It seems the more intuitive method would be by substitution and working the +x and -x branches as separate integrals. Curiously, if he hadn't stopped at 8:14 but repeated the same mistake by re-writing the final answer as (√(x²-1)) / x he would accidentally found the correct antiderivative. Even so, the fact that the function is undefined on [-1,1] should lead to caution in computing definite integrals.
Integration by parts solves the problem but you avoid it with all costs x^4+1 = (1-x^4)+2x^4 and separate integral into two integrals then integral -\int{\frac{\sqrt{x^4-1}}{x^2}}dx calculate by parts with u = \sqrt{x^4-1} , dv = -\frac{1}{x^2}dx Answer is \frac{\sqrt{x^4-1}}{x}+C
It is good example for practising integration by parts especially if we teach basics of integral calculus No other techniques necessary so it is good example for teaching integration by parts
Looks clever, but your comment isn't clear or complete. Most normal people learning calculus would not think to rewrite the numerator like that or do integration by parts. It's interesting that if Mr. Calculus Booster had repeated his mistake again at the end, to change his answer into your result, it would accidentally give correct definite integral results.
2:10 Wrong.
There is nothing in the question to suggest that x must be positive, so √(x²) = |x| , not x.
Looking at the graph of the integrand, it is positive everywhere it's defined. So any interval where it's defined will have a positive definite integral. But if we try to use the expression boxed at 8:14 to calculate the area under the curve from (for example) -20 to -10 we get a (wrongly) negative answer.
This video is a good example of what can go wrong when not thinking clearly about roots of powers!
It seems the more intuitive method would be by substitution and working the +x and -x branches as separate integrals.
Curiously, if he hadn't stopped at 8:14 but repeated the same mistake by re-writing the final answer as (√(x²-1)) / x he would accidentally found the correct antiderivative. Even so, the fact that the function is undefined on [-1,1] should lead to caution in computing definite integrals.
Excellent Point! I thought the same thing.
gt
Integration by parts solves the problem
but you avoid it with all costs
x^4+1 = (1-x^4)+2x^4
and separate integral into two integrals
then integral -\int{\frac{\sqrt{x^4-1}}{x^2}}dx
calculate by parts with u = \sqrt{x^4-1} , dv = -\frac{1}{x^2}dx
Answer is \frac{\sqrt{x^4-1}}{x}+C
It is good example for practising integration by parts especially if we teach basics of integral calculus
No other techniques necessary so it is good example for teaching integration by parts
Looks clever, but your comment isn't clear or complete.
Most normal people learning calculus would not think to rewrite the numerator like that or do integration by parts.
It's interesting that if Mr. Calculus Booster had repeated his mistake again at the end, to change his answer into your result, it would accidentally give correct definite integral results.
（(x^4 ＋ 1)/(2x sqrt(x^4 - 1））+ c