Dilemma of Integrals

Calculus Level 3

Which of the following statements regarding the functions $f(x),g(x)$ for all real values $x$ are not true?

If $\displaystyle\int_{a}^{b}f(x)dx=\displaystyle\int_{a}^{b}g(x)dx$ and $f(b)=g(b), f(a)=g(a)$ , then $f(x)\equiv g(x)$ .
If $\displaystyle\int_{a}^{b}f(x)dx\geq \displaystyle\int_{a}^{b}g(x)dx$ , then $f(b)\geq g(b)$ and $f(a)\geq g(a)$ .
$\displaystyle\int_{0}^{b}f(x)dx=\displaystyle\int_{a}^{0}f(x)dx$ and $b+a=0$ if and only if $f(x)$ is an even function.
$\displaystyle\int_{0}^{b}g(x)dx=\displaystyle\int_{0}^{a}g(x)dx$ and $b-a=0$ if and only if $g(x)$ is an odd function.

You can try more of my fundamental problems here .

1,2,3,4 only 1,2,4 only 1,2,3 only 2,3,4 only

2 solutions

Jc 506881
Jan 29, 2018

Counterexamples to each of the statements above can obtained using the same idea: Just because two regions have the same area doesn't mean they have the same shape.

Donglin Loo
Jan 30, 2018

We shall provide counter-examples to prove the statements wrong.

In Statement 1,

Let $f(x)=\sin x, g(x)=\cos x$

$\displaystyle\int_{0}^{2 \pi}sinxdx=0,\displaystyle\int_{0}^{2 \pi}cosxdx=0$

$\displaystyle\int_{0}^{2 \pi}f(x)dx=\displaystyle\int_{0}^{2 \pi}g(x)dx=0$

$f(x)\neq g(x)$

So, Statement 1 is false.

In Statement 2,

Let $f(x)=\sin x, g(x)=\cos x$

$\displaystyle\int_{0}^{\cfrac{\pi}{2}}sinxdx=\displaystyle\int_{0}^{\cfrac{\pi}{2}}cosxdx$

$f(0)=0, g(0)=1$ $\Rightarrow f(0)<g(0)$

So, Statement 2 is false.

In Statement 3 and Statement 4,

We shall suppose that $a=b=0$

For all real-valued functions $p(x)$ , $\displaystyle\int_{0}^{0}p(x)dx=0$ , regardless of $p(x)$ is even, odd or neither.

So, Statement 3 and Statement 4 are false.

$\textbf{Note:}$ : The term $\textbf{A if and only if B}$ means that if $\textbf{A}$ , then $\textbf{B}$ and also if $\textbf{B}$ , then $\textbf{A}$ . It must work in both ways.

Dilemma of Integrals

2 solutions

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