"Evaluate the definite integral. ∫01xe−x2dx" was answered by our community

gainejavima

Answered question

2021-11-22

Evaluate the definite integral.

\int_{0}^{1} x e^{- x^{2}} d x

Harr1957

Beginner2021-11-23Added 18 answers

Evaluate the definite integral.

\int_{0}^{1} x e^{- x^{2}} d x

Let

x^{2} = t

2x dx=dt

x d x = \frac{d t}{2}

\int_{0}^{1} x e^{- x^{2}} d x = \int_{0}^{1} e^{- t} \frac{d t}{2}

= \frac{1}{2} \int_{0}^{1} e^{- t} d t

= \frac{- 1}{2} {[e^{- t}]}_{0}^{1}

= \frac{- 1}{2} [e^{- 1} - 1]

Warajected53

Beginner2021-11-24Added 12 answers

Step 1: Remove parentheses.

\int_{0}^{1} x e^{- x^{2}} d x

Step 2: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:

\int_{a}^{b} f (x) d x = F (x) ∣_{a}^{b} = F (b) - F (a)

Step 3: In this case,

f (x) = x e^{- x^{2}}

. Find its integral.
Step 4: Since

F (x) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(1)−F(0):

- \frac{e^{- 1^{2}}}{2} - (- \frac{e^{- 0^{2}}}{2})

Step 5: Simplify.

- \frac{e^{- 1}}{2} + \frac{e^{- 0}}{2}

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