Evaluate [0, ∞]∫𝑒^−5𝑥^2𝑑𝑥

Answer 1

well, this is a special integral. You have to know that
[0,∞]∫e^−x^2 dx = √π/2
Now substitute u = √5 x and get
[0,∞]∫e^−5x^2 dx = √(π/5)/2

for more info, read up on erf(x)

Answer 2

I don't understand something. Why does the book answer is π/2√5 but your √(π/5)/2. Do you know how?

Answer 3

come on -- this is Algebra I
√(π/5)/2 = (√π/√5)/2 = √π / 2√5

(a/b)/c = a/(bc)

Answer 4

Your answer still did match the book answer. The book answer again π/2√5 but your √π / 2√5. You need to cancel the squreroot but how?

Answer 5

I guess your book has a typo. See

https://www.wolframalpha.com/input/?i=+%E2%88%AB%5B0+..+%E2%88%9E%5De%5E%28%E2%88%925x%5E2%29+dx