MHB How to prove this corollary in Line Integral using Riemann integral

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To prove the corollary in line integrals using the Riemann integral, consider a smooth curve C with arc length L and a vector field f(x, y) bounded by M. The integral of f along C can be expressed as the integral of f evaluated at a parametrization r(t) over the interval [a, b]. By applying the hint that the absolute value of an integral is less than or equal to the integral of the absolute value, it follows that the integral is bounded by the product of M and the integral of the derivative of the parametrization, |r'(t)|. This leads to the conclusion that the absolute value of the line integral is less than or equal to ML, confirming the corollary.
WMDhamnekar
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. Let C be a smooth curve with arc length L, and suppose that f(x, y) = P(x, y)i +Q(x, y)j is a vector field such that $|| f|(x,y) || \leq M $ for all (x,y) on C. Show that $\left\vert\displaystyle\int_C f \cdot dr \right\vert \leq ML $
Hint: Recall that $\left\vert\displaystyle\int_a^b g(x) dx\right\vert \leq \displaystyle\int_a^b |g(x) | dx $ for Riemann Integrals.

Now, how to prove this corollary using this hint?
 
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Assuming $C$ is parametrized by $r(t)$, $a\le t \le b$, write $$\int_C f\cdot dr = \int_a^b f(r(t))\cdot r'(t)\, dt$$ By the hint, $\int_C f\cdot dr$ is bounded by $\int_a^b |f(r(t))\cdot r'(t)|\, dt$. Since $|f(r(t))\cdot r'(t)| \le |f(r(t))| |r'(t)| \le M |r'(t)|$ for all $t$, we deduce $\left|\int_C f\cdot dr\right| \le M \int_a^b |r'(t)|\, dt = ML$.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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