Exercise 8.4.12

Question

Assume the function $f(x,t)$ is continuous on the rectangle $D = \{(x,t):a \leq x \leq b,c \leq t \leq d\}$. Explain why the function
$$F(x) = \int_c^d f(x,t) dt$$
is properly defined for all $x \in [a,b]$.

Ulisse Mini · Accepted Answer

For a fixed $x$, $f(x,t)$ is continuous with respect to $t$. To see this, we need to show $\forall t_0 \in [c,d],\epsilon > 0$, we can find $\delta$ so that $\left|t - t_0\right| < \delta $ implies $\left|f(x,t) - f(x, t_0)\right| < \epsilon$. But since
$$\left|t - t_0\right| = \|(x,t) - (x, t_0) \|$$
we can just use the continuity of $f$ over $D$ to conclude $f(x,t)$ is continuous with respect to $t$, and therefore $\int_c^d f(x,t) dt$ is properly defined.

Exercise 8.4.12

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Exercise 8.4.12

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