A minimum value of int_0^x t e^{t^2} d t is | vedclass

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(A) Let $f(x) = \int_0^x t e^{t^2} dt$. To evaluate the integral, substitute $u = t^2$, so $du = 2t dt$, which implies $t dt = \frac{1}{2} du$. When $

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(A) Let $f(x) = \int_0^x t e^{t^2} dt$. To evaluate the integral, substitute $u = t^2$, so $du = 2t dt$, which implies $t dt = \frac{1}{2} du$. When $t = 0$, $u = 0$. When $t = x$, $u = x^2$. Thus, $f(x) = \int_0^{x^2} \frac{1}{2} e^u du = \frac{1}{2} [e^u]_0^{x^2} = \frac{1}{2} (e^{x^2} - 1)$. To find the minimum, differentiate $f(x)$ with respect to $x$: $f'(x) = \frac{1}{2} (e^{x^2} \cdot 2x) = x e^{x^2}$. Setting $f'(x) = 0$ gives $x = 0$. Now, find the second derivative: $f''(x) = e^{x^2} + x(e^{x^2} \cdot 2x) = e^{x^2} (1 + 2x^2)$. At $x = 0$, $f''(0) = e^0 (1 + 0) = 1$. Since $f''(0) > 0$, the function has a local minimum at $x = 0$. The minimum value is $f(0) = \frac{1}{2} (e^0 - 1) = \frac{1}{2} (1 - 1) = 0$.

$A$ minimum value of $\int_0^x t e^{t^2} d t$ is

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