To find the line integral of [tex]\( f(x, y) = y e^{x^2} \)[/tex] along the curve [tex]\( r(t) = -4t \mathbf{i} + 3t \mathbf{j} \)[/tex] for [tex]\( -2 \leq t \leq -1 \)[/tex], follow these steps:
### Step 1: Parameterize the Curve
Given the parameterization [tex]\( r(t) \)[/tex]:
- [tex]\( x(t) = -4t \)[/tex]
- [tex]\( y(t) = 3t \)[/tex]
### Step 2: Compute [tex]\( dx \)[/tex] and [tex]\( dy \)[/tex]
Differentiate [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
- [tex]\( \frac{dx}{dt} = \frac{d}{dt}(-4t) = -4 \)[/tex]
- [tex]\( \frac{dy}{dt} = \frac{d}{dt}(3t) = 3 \)[/tex]
### Step 3: Express [tex]\( f(x, y) \)[/tex] in Terms of [tex]\( t \)[/tex]
Substitute [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex] into [tex]\( f(x, y) \)[/tex]:
[tex]\[
f(x(t), y(t)) = y(t) e^{(x(t))^2} = 3t e^{(-4t)^2} = 3t e^{16t^2}
\][/tex]
### Step 4: Formulate the Line Integral
The line integral of [tex]\( f(x, y) \)[/tex] along the curve is given by:
[tex]\[
\int_{a}^{b} f(x(t), y(t)) \left( \frac{dx}{dt} dt \right)
\][/tex]
Here, [tex]\( a = -2 \)[/tex] and [tex]\( b = -1 \)[/tex].
Compute:
[tex]\[
\frac{dx}{dt} = -4
\][/tex]
Substitute into the integral:
[tex]\[
\int_{-2}^{-1} 3t e^{16t^2} (-4) dt = -12 \int_{-2}^{-1} t e^{16t^2} dt
\][/tex]
### Step 5: Evaluate the Integral
To evaluate the integral [tex]\( -12 \int_{-2}^{-1} t e^{16t^2} dt \)[/tex]:
Use substitution [tex]\( u = 16t^2 \)[/tex], then [tex]\( du = 32t dt \)[/tex] or [tex]\( \frac{du}{32} = t dt \)[/tex].
Change the limits of integration with respect to [tex]\( u \)[/tex]:
- When [tex]\( t = -2 \)[/tex], [tex]\( u = 16(-2)^2 = 64 \)[/tex]
- When [tex]\( t = -1 \)[/tex], [tex]\( u = 16(-1)^2 = 16 \)[/tex]
Thus, the integral becomes:
[tex]\[
-12 \int_{64}^{16} e^u \frac{du}{32} = -\frac{12}{32} \int_{64}^{16} e^u du = -\frac{3}{8} \int_{64}^{16} e^u du
\][/tex]
### Step 6: Solve the Integral
[tex]\[
-\frac{3}{8} \int_{64}^{16} e^u du = -\frac{3}{8} \left[ e^u \right]_{64}^{16}
\][/tex]
Evaluate the definite integral:
[tex]\[
-\frac{3}{8} \left( e^{16} - e^{64} \right) = -\frac{3}{8} e^{16} + \frac{3}{8} e^{64}
\][/tex]
### Step 7: Write the Final Answer
The value of the line integral is:
[tex]\[
= \frac{3}{8} e^{64} - \frac{3}{8} e^{16}
\][/tex]
Simplifying, we get:
[tex]\[
= -3 \frac{e^{16}}{8} + 3 \frac{e^{64}}{8}
\][/tex]
Thus, the final result of the line integral is:
[tex]\[
-\frac{3}{8} e^{16} + \frac{3}{8} e^{64}
\][/tex]
Therefore, the line integral of [tex]\( f(x, y) \)[/tex] along the curve [tex]\( r(t) \)[/tex] from [tex]\( t = -2 \)[/tex] to [tex]\( t = -1 \)[/tex] is:
[tex]\[
-3 \frac{e^{16}}{8} + 3 \frac{e^{64}}{8}
\][/tex]
