${ message }

Your cart is empty

Discount (${discount_percentage}%) : - ${discount}KD

Need Help?

How was the speed of the video ?

How well did you understood the video ?

Was the video helpful?

Was the notes helpful?

Sign up to try our free practice

KD

5.850

1 month

70% off

19.500

Add to cart

11.850

4 months

39.500

Subscribe to Calculus B

Practice (Free)

Practice

Find the area of the region bounded by the curves \(y=\sin (x), y=\cos (x), x=0, x=\frac{\pi}{2}\)

\(A \equiv A_{1}+A_{2}\)

\(=\int_{0}^{\frac{\pi}{4}}(\cos (x)-\sin (x)) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin (x)-\cos x) d x \)

\(=[\sin (x)+\cos (x)]_{0}^{\frac{\pi}{4}}+[-\cos (x)-\sin (x) ]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \)

\(=\left[\sin \frac{\pi}{4}+\cos \frac{\pi}{4}-(\sin (0)+\cos (0))\right] \)

\(+\left[-\cos \frac{\pi}{2}-\sin \frac{\pi}{2}-\left(-\cos \frac{\pi}{4}-\sin \frac{\pi}{4}\right)\right] \)

\(=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-(0+1)\right]+\left[-0-1-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\right] \)

\(=\left[\frac{2}{\sqrt{2}}-1\right]+\left[-1+\frac{2}{\sqrt{2}}\right]=\frac{2}{\sqrt{2}}-1-1+\frac{2}{\sqrt{2}}=\frac{4}{\sqrt{2}}-2=2 \sqrt{2}-2 \)

No comments yet

Find the area of the region bounded by the curves

\(y=\sin (x), y=\cos (x), x=0, x=\frac{\pi}{2}\)

\(A \equiv A_{1}+A_{2}\)

\(=\int_{0}^{\frac{\pi}{4}}(\cos (x)-\sin (x)) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin (x)-\cos x) d x \)

\(=[\sin (x)+\cos (x)]_{0}^{\frac{\pi}{4}}+[-\cos (x)-\sin (x) ]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \)

\(=\left[\sin \frac{\pi}{4}+\cos \frac{\pi}{4}-(\sin (0)+\cos (0))\right] \)

\(+\left[-\cos \frac{\pi}{2}-\sin \frac{\pi}{2}-\left(-\cos \frac{\pi}{4}-\sin \frac{\pi}{4}\right)\right] \)

\(=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-(0+1)\right]+\left[-0-1-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\right] \)

\(=\left[\frac{2}{\sqrt{2}}-1\right]+\left[-1+\frac{2}{\sqrt{2}}\right]=\frac{2}{\sqrt{2}}-1-1+\frac{2}{\sqrt{2}}=\frac{4}{\sqrt{2}}-2=2 \sqrt{2}-2 \)

No comments yet

## Join the conversation