Let , $f:\mathbb R\to \mathbb R$ be a differentiable function such that $f'$ is bounded. Given a closed and bounded interval $[a,b]$ and partition $P=\{a=a_0<a_1<\cdots <a_n=b\}$ of $[a,b]$ . Let , $M(P,f)$ and $m(P,f)$ denotes the upper Riemann sum and lower Riemann sum respectively. Then which are correct ?
(A) $$\left|M(P,f)-\int_a^bf(x)\,dx\right|\le (b-a).\sup\{|f(x)|:x\in[a,b]\}$$
(B) $$\left|m(P,f)-\int_a^bf(x)\,dx\right|\le (b-a).\inf\{|f(x)|:x\in[a,b]\}$$
(C) $$\left|M(P,f)-\int_a^bf(x)\,dx\right|\le (b-a)^2.\sup\{|f'(x)|:x\in[a,b]\}$$
(D) $$\left|m(P,f)-\int_a^bf(x)\,dx\right|\le (b-a)^2.\inf\{|f'(x)|:x\in[a,b]\}$$
Attempt :
For option (A) , $$\left|M(P,f)-\int_a^bf(x)\,dx\right|\le |M(P,f)|+\left|\int_a^bf(x)\,dx\right|\le 2M(b-a)$$As , $M_r\le M$ , so $M(P,f)\le M(b-a)$ ; where , $M=\sup_{a\le x\le b} f(x)$ and $M_r=\sup_{x_{r-1}\le x \le x_r}f(x)$.
Where my mistake ? How I can prove or disprove ? Please suggest..
(A) is false: On $[0,1]$ take $f(x)= x^2-1/2$ and $P= \{0,1\}.$ Then $M(P,f) = 1/2, \int_0^1f = -1/6, \sup |f| = 1/2.$ So the left side equals $2/3,$ the right side equals $1/2.$
(C ) is true: We have
$$M(P,f) - \int_a^b f = \sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}(M_k - f) = \sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}(f(c_k) - f(t))\ dt$$ $$ \implies |M(P,f) - \int_a^b f| \le \sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}|f(c_k) - f(t)|\ dt.$$
The $c_k$ arise from the fact that $f$ is continuous, hence achieves a maximum on each subinterval. Let $C= \sup |f'|.$ By the MVT, $|f(c_k) - f(t)|\le C(x_k-x_{k-1}) \le C(b-a).$ So the last sum above is $\le$
$$ C(b-a)\sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}dt = C(b-a)^2 .$$