I know what the sine function is, and why its graph is a wave. I know what a definite integral is. But I don't understand why this integral happens to be $1$. Can anyone explain it to me?

Note: Someone tried to edit this saying that I know how to evaluate definite integrals. I don't. I know what they are though.

  • $\begingroup$ If evaluating the integral with the fundamental theorem of calculus does not give you insight, you should think about why the fundamental theorem of calculus is true. $\endgroup$ – M. Van Jun 5 '16 at 18:21
  • $\begingroup$ We have $f(x)= \sin x$. In order to integrate, you need to find the antiderivate $F$ which is $-\cos x$ and then just fill in the upper and lower limits of your integral. In general: $\int_a^{b} f(x)\, dx= F(b) - F(a)$ $\endgroup$ – Tesla Jun 5 '16 at 18:27
  • $\begingroup$ Actually, I couldn't agree more. (That goes for both comments, but was meant for the first.) $\endgroup$ – asher drummond Jun 5 '16 at 18:28
  • $\begingroup$ I wonder what the reason for the downvoter downvoting the answers and the question itself. I see that the question, sigma's answer, and my answer have all received downvotes. $\endgroup$ – user342209 Jun 5 '16 at 18:39
  • $\begingroup$ Although I didn't down vote I feel like a Riemann sum would be more helpful in this circumstance . @user342209 $\endgroup$ – Ahmed S. Attaalla Jun 5 '16 at 18:47

$\int_{0}^\frac{\pi}{2}\sin x\,dx=-\cos x\mid_0^\frac{\pi}{2}=0 -(-1)=1$

$F(x)\mid_a^{b}$ means $F(b)-F(a)$

So in general $\int_a^{b} f(x)\, dx= F(b) - F(a)$


The value $I$ $$ I = \int\limits_0^{\pi/2}\sin x \,dx = 1 $$ can be interpreted as area between the graph of $\sin$ and the $x$-axis.

$$ A = \int\limits_0^{\pi/2} \lvert \sin x \rvert \,dx = \int\limits_0^{\pi/2}\sin x \,dx = I $$

See the image below which uses a unit raster of $0.2 = 1/5$, so a unit square has area $1/25$. You will see that roughly 25 units fit the area.

Note the excess area to the right of $x=1.5$ of about $5\times 0.1=0.5$.

area interpretation


$$\int_0^{\frac{\pi}{2}}\sin(x)dx=-\cos(x)\large|^{\pi/2}_0$$ Can you take it from here?

In case you mean to ask what the 'meaning' of the Integral is, the Integral simply computes the area between the graphs of $\sin(x)$ and the $X$ axis between $x=0$ and $x=\frac{\pi}{2}$

  • $\begingroup$ I can't give any better answer than this, but I suspect that the OP wants something else (assuming they do, in fact, understand definite integrals.) $\endgroup$ – Semiclassical Jun 5 '16 at 18:21
  • $\begingroup$ Perhaps the OP wanted to know what the Integral actually represents? $\endgroup$ – user342209 Jun 5 '16 at 18:23
  • $\begingroup$ Actually, this works for me, but I have to wait 9 minutes (at the time of this comment being posted) to accept this answer. $\endgroup$ – asher drummond Jun 5 '16 at 18:23
  • $\begingroup$ Read the above comment. $\endgroup$ – asher drummond Jun 5 '16 at 18:25
  • 1
    $\begingroup$ $$\int_a^bf(x)dx=F(x)|_a^b=F(b)-F(a)$$ when the function $F(x)$ is defined at both $a$ and $b$. In case they are not defined, you have to take limits depending on the Interval of Integration. Note that $F(x)$ is the antiderivative of $f(x)$. $\endgroup$ – user342209 Jun 5 '16 at 18:36

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