Question

guess an antiderivative for the integrand function. validate your guess by differentiation and then evaluate the given definite integral (hint: keep in mind the chain rule in guessing an antiderivative.)
(int_{0}^{6} xe^{x^2} , dx)

(int_{0}^{6} xe^{x^2} , dx = square)
(type an exact answer in terms of ( e ).)

Explanation:

Step1: Guess the antiderivative

Let's consider the function \( f(x) = xe^{x^2} \). We know that the derivative of \( e^{u} \) with respect to \( x \) is \( e^{u} \cdot u' \) by the chain rule. Let \( u = x^2 \), then \( u' = 2x \). Notice that our integrand has \( x \), which is half of \( 2x \). So we guess the antiderivative \( F(x)=\frac{1}{2}e^{x^2} \).

Step2: Validate by differentiation

Differentiate \( F(x)=\frac{1}{2}e^{x^2} \) with respect to \( x \). Using the chain rule, \( F'(x)=\frac{1}{2}\cdot e^{x^2}\cdot 2x = xe^{x^2} \), which matches the integrand. So our guess is correct.

Step3: Evaluate the definite integral

Using the Fundamental Theorem of Calculus, \( \int_{a}^{b}f(x)dx = F(b)-F(a) \). Here, \( a = 0 \), \( b = 6 \), and \( F(x)=\frac{1}{2}e^{x^2} \). So we calculate \( F(6)-F(0) \).

\( F(6)=\frac{1}{2}e^{6^2}=\frac{1}{2}e^{36} \)

\( F(0)=\frac{1}{2}e^{0^2}=\frac{1}{2}e^{0}=\frac{1}{2}\cdot 1=\frac{1}{2} \)

Then \( \int_{0}^{6}xe^{x^2}dx = F(6)-F(0)=\frac{1}{2}e^{36}-\frac{1}{2}=\frac{e^{36}-1}{2} \)

Answer:

\(\frac{e^{36}-1}{2}\)