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# Find the area of the shaded region.

## $\frac{1}{2}(e-1)$

#### Topics

Applications of Integration

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### Video Transcript

right. This problem. We need to find area of the shaded regions. So buy formula. We know. Ah, if we want to inter grow instead ofthe X first of the red on the integral. So the area Hey, no e coast too into grow in terms ofthe X And we know the shaded area X goes from zero to one so we can write the bound it pounds from zero to one. Okay. Ah, for here. Ah, the formula is upper curve minus the lower occur. So this will be you two The power X linus x times e to the power X square. So our next step basically is Try toe Aah! Evaluate this, integral. All right, So we can do it. Except to separately the first term. Well, the ranger grow from zero to one into the power ext the ex minus integral from zero to one times to the power X square, the ex. Okay. And the first that we know the anti directly about that. This is e to the Power X. So this is such a yeah to the power X evaluate acts because one minus X equals zero. Okay. And this term here we can bring this acts to the differential that will give us X square. So before we do that, we need two times two. So what we trick they're using here every times two and it divided by two. So the second term will be behalf ties into grove the roads one you two the power X square King X square. And as we can see this if we use change of arable say this explorer to be another variable t Then this is you to the party. So the auntie do it Who will be itself. So this term Ah, you free right here. No B negative One half times into the power X square evaluated and ex ecos one minus x Go to zero. Right? So, in other words, this is, um So next page the first term, remember, is need to the power axe. You're only right from one minus X equals in Cyril zero um, minus one half he to the Power X square. Valerie had taxi. Could one minus X equals zero. So no and serve. It will be the first term. Here is I e to the power of one. Linus Ito, the power zero which is what and the second term and second term is minus one half times the first of eyes that you two, the power one. So it's just e to the tower one minus into a powered zero where she's far and as you can see, answer will be one half times minus one, right.

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Applications of Integration

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