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## $\frac{13}{5}$

#### Topics

Applications of Integration

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##### Top Calculus 2 / BC Educators  ##### Heather Z.

Oregon State University  Lectures

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### Video Transcript

sketch the region enclosed by the given curves and find its area. So we have two functions. Why equals X to the fourth and why equals two minus the absolute value of X on dhe, they are graft over here in green and blue. On the right hand side, when we want to find area and we're trying to construct are integral To find the area that's enclosed between these two graphs, we need to also consider our boundary points. So we need to look for those points of intersection where the two graphs intersect and from the X Y tables we can see that at X equals negative one y equals one and X equals one y equals one. We have points of intersections where these two graphs indeed intersect each other. Now what's so beautiful about these graphs is that they are symmetric. And what's interesting about the blue graph is that this is in fact, a piece wise function. The absolute value of X is so if we split the blue graft down in half by the Y axis on the left hand side, we get the function F of X equals X plus two and on the right hand side. We get the function G of X equals negative X plus two. And this is helpful because since there is a cemetery involved in these graphs, there's also cemetery involved in the region enclosed by these graphs. So, in fact, we can just calculate the region of half, and that region will call it our, since he's on the right hand side is the same area as the region on the left. So all I need are the boundaries for the right hand side, and I can create an integral from that, multiply it by two, and I'll get the area for the entire region. So the boundary, the lower bound for the right region, would be X zero and then and that's because of the Y axis. That's where we're starting. And then for the right hand side, the upper bound of the exes one. And that's because of the point of intersection there. That's the X value. So in constructing the integral, we would have to if my panel work there, we go to times Thean, a girl from the lower bound zero to the upper bound one, and then we went to subtract the lower function. So the green function from the upper function from the blue function. So we want G of X minus hour axe to the fourth so g of X is negative X plus two and subtract X to the fourth from that. And that's where the respect to X so d of X dx and then we have two times negative 1/2 x squared plus two x minus 1/5 x to the fifth, all evaluated from 0 to 1 That gives us two times. Now we plug in the upper bound first and then we subtract the value resulting from plugging in the lower bound. So two times negative 1/2 plus two. No, fix that. Here we go, minus 1/5 that minus plugging and zero. Well, that all goes to zero. So we have to times we want a common denominator here. So we have negative five over 10 plus 20 over 10 minus two over 10. That gives us two times 13 over 10 which two in 10 cancel to visit to once it wasn't attend five times. So we have the area between those two curves. Is 13 a year five. That's the total area between the blue curve and the green kerf Missouri State University

#### Topics

Applications of Integration

##### Top Calculus 2 / BC Educators  ##### Heather Z.

Oregon State University  Lectures

Join Bootcamp