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SL

# Find the area of the shaded region.

## $A=9$

#### Topics

Applications of Integration

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### Video Transcript

okay. As we see, this problem is represented all the function representing back why on So when they tried to evaluate the interviewer off the shaded region of Integral in terms ofthe why instead of X, um so a region a shader region, he coast too integral in terms ofthe why y in the bound for y ears go. It goes from zero to three. Here is the three here He said origin zero, they go. Well, it goes from zero to three. Ah, and angel voices just up her curved minus the lower curved upper curve. Here is this section which is to y minus y square. So why minus y square? This is the upper curve minus the lower curve is Ritchie's. Why square minus for Why? Okay, so why square liners for what? Um and I ve Kamp Buddha by square together and the y terms together. So this will be zero to three on the double. Minus will be. Plus, So it's six. Why? Minus to y square. You are. And the anti directive for this is, um, three y square, minus two third like you. You ready? And that had the bound Richie Savaiko two three minus like Oh two zero! Ah, and we know when while you go to zero, this whole thing is a zero. So this whole Yuko's Tio with Plug in by three, this will be three times. Ni reaches twenty seven Linus to third times three Q b is twenty seven. So twenty seven. So the answer will be on to third times twenty seven. This is eighteen. Oh, and there will be nice.

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#### Topics

Applications of Integration

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