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For the inequality $\sin x \le x$ which holds for $x \ge 0$, Find the upper bound for the value of $\int_0^1 \sin x dx$, express it in a simplified fraction or an integer.

Domination rule states that $\int_a^b f(x) dx \ge \int_a^b g(x) dx$ if $f(x) \ge g(x)$ on $[a,b]$.

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

Its clear that for the interval $[0,1]$, the "area under graph" for the function $f(x)$ will be larger than $g(x)$.

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What is the meaning of "upper bound"?

Is it just taking area under graph of $f(x)$ minus that of $g(x)$? But this doesn't give me a simplified fraction because $\sin 1$ is always going to give a decimal answer.

user307640
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1 Answers1

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Upper bound is the value which a function is never going to cross.

For example, upper bound for $\sin x$ is $1$ as its value is never going to be bigger than $1$.

We can also say upper bound for $\sin x$ is $2$ because its value is never going to cross $2$. (It's not going to cross even $1$, that's separate.)

We may say that the lowest upper bound for $\sin x$ is $1$. Also, that value is attained at $\frac\pi2$, that means $\frac{3.14}2=1.57$.

But the given interval is $0$ to $1$. Here, we know that upper bound for $\sin x$ is $x$. (It's not the lowest upper bound, that's separate).

Thus, $$\int_0^1\sin xdx\le\int_0^1xdx=\frac12$$

aarbee
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