Question

Find the area enclosed by the graphs of y = sin x, y = cos x, x = 0, and x = 2. Sketch a graph.

Answer 1

Answer:

1.34 (3 sf)

Step-by-step explanation:

To find the area between 2 curves:

$\int\limits^a_b {(top-bottom)} \, dx$

As these 2 curves intersect, we have to first calculate where they intersect, then work out the area between the curves BEFORE and AFTER the point of intersection, as the "top" and "bottom" curves change after the point of intersection (see attached graph).

   sin(x) = cos(x)

⇒ tan(x) = 1

⇒ x = $\frac{1}{4} \pi$

So the curves intersect at $x=\frac{1}{4} \pi$ (between x = 0 and x = 2)

To calculate the area between the 2 curves, $x=0$ and $x=\frac{1}{4} \pi$:

$\int\limits^a_b {(top-bottom)} \, dx=\int\limits^\frac{\pi}{4}_0 {(cos(x)-sin(x))} \, dx=\sqrt{2}-1$

Then to calculate the area between the 2 curves, $x=\frac{1}{4} \pi$ and $x=2$:

$\int\limits^a_b {(top-bottom)} \, dx=\int\limits^2_\frac{\pi}{4} {(sin(x)-cos(x))} \, dx=0.92106...$

Therefore, the total area = $\sqrt{2} -1+0.92106...=1.335276534...$

(Please Brainliest if helped - thanks!)