Sketch the region enclosed by the given curves and find its...

Key Concepts

Evaluation of Definite Integrals

Evaluating definite integrals involves the computation of the accumulated area between the curves over the specified interval. This step uses the Fundamental Theorem of Calculus to calculate the exact area of the region by finding antiderivatives and applying the integration limits.

Setting Up Definite Integrals for Area

Once the region is clearly defined through curve sketching and intersection points, the next step is to set up the appropriate definite integral. This involves expressing the area as an integral with respect to a suitable variable, carefully establishing the limits of integration based on the intersection points and ensuring the integrand represents the difference between the upper and lower bounding functions.

Finding Intersection Points

Finding the points of intersection between curves is essential in identifying the limits of integration. By solving the equations of the curves simultaneously, one can determine where the curves cross each other, which defines the boundary of the region whose area is to be calculated.

Curve Sketching

Curve sketching involves drawing a rough graph of the functions to understand their behavior, such as the shape, intercepts, turning points, and asymptotic behavior. This helps in visualizing the region enclosed by the curves and determining which function forms the upper or lower boundary of the integration region.

Transcript

00:02 Okay, sketch the region enclosed by the given curves and find its area.

00:07 So first let's draw those two curves.

00:10 This is our xy plane.

00:13 And the first function is x equals to y to the power 4.

00:18 So it looks like a problem, but it grows much faster.

00:23 So it opens to the right like this.

00:30 This will be y x equals to y to the power 4.

00:37 And then the second term here, y equals to square root of 2 minus x.

00:50 This, we know it can be represented as if we take the square, this is just y square equals to 2 minus x.

01:04 And we move around things.

01:07 This is equivalent with x equals to 2.

01:16 Minus y square yeah and be careful here we also need this indication why is greater or equal to zero because we know the square root by equal to the square root of something so why cannot be negative okay and this is also a problem it opened to the left it started at 2 and open to the left so the in close, the area will be here.

02:05 And here is 2, here is 0.

02:10 And as we can see, since we format the second curve, right now everything is represented by y.

02:26 So if we want to find the area, take the integral with respect to y instead of x.

02:35 And the boundary is this point and this point.

02:45 So if i deno this as 1, this is y, 2.

02:51 How to find them? it's just, oh, be careful here.

03:01 The enclosed region is actually now the entire thing because we got this extra condition here when we reformulate the second curve.

03:19 Y should be non -negative.

03:23 So this enclosed area is only this half of them.

03:33 Here is empty, okay? therefore, we don't have this y1...