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Let π be a function given by π of π₯ is equal to the integral between zero and π₯ of negative π to the π‘ squared with respect to π‘.
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Which of the following must be true on the interval where π₯ is greater than zero but less than two. a) π of π₯ is greater than zero, decreasing, and concave up. b) π of π₯ is less than zero, decreasing, and concave down. c) π of π₯ is less than zero, decreasing, and concave up.
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And d) π of π₯ is greater than zero, decreasing, and concave down.
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Letβs see if we can find any further information about π of π₯ and negative π to the π‘ squared.
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In order to do this, we will be using the fundamental theorem of calculus, which tells us that the integral between π and π of π prime of π‘ with respect to π‘ is equal to π of π minus π of π.
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If we let negative π to the π‘ squared be equal to π prime of π‘ and π equal to zero and π equal to π₯, then we can say that the integral between zero and π₯ of π prime of π‘ with respect to π‘ is equal to π of π₯ minus π of zero.
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However, this must also be equal to our integral from the question since we defined π prime of π‘ as negative π to the π‘ squared.
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And we know that our integral in the question is also equal to π of π₯.
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And so here, we have formed a relationship between π of π₯ and π of π₯.
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We have that π of π₯ minus π of nought is equal to π of π₯.
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We can differentiate both sides of this equation with respect to π₯ since π of nought is a constant that will go to zero.
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And weβre left with π prime of π₯ is equal to π prime of π₯.
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However, weβve defined π prime of π‘ as negative π to the π‘ squared.
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Therefore, we can say that π prime of π₯ is equal to negative π to the π₯ squared.
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Now π prime of π₯ is the derivative of π with respect to π₯, which tells us the slope of π.
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Now, π to the π₯ squared is an exponential.
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And we will have a positive value for any value of π₯.
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Therefore, negative π to the π₯ squared must be negative for all values of π₯.
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This means that the slope of π is negative for all values of π₯.
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So we can say that π is decreasing.
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Now that we found the first derivative of π, letβs consider the second derivative of π since this will tell us about the concavity of π.
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Second derivative of π, also known as π double prime of π₯, itβs equal to the derivative of π prime of π₯ with respect to π₯.
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And we know that π prime of π₯ is equal to negative π to the π₯ squared.
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So we simply need to differentiate this function.
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Now, this is a compound function.
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We have a function of π₯ squared within an exponential.
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Therefore, we must use the chain rule in order to differentiate it.
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The chain rule tells us that dβ by dπ₯ is equal to dβ by dπ’ times dπ’ by dπ₯.
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In our case, we can let β be equal to negative π to the π’ and π’ be equal to π₯ squared.
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Therefore, when we combine these two functions, we will arrive back at our original function.
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Applying the chain rule, we obtained that d by dπ₯ of negative π to the π₯ squared is equal to d by dπ’ of negative π to the π’ times d by dπ₯ of π₯ squared.
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The first term here is exponential.
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So when we differentiate it, nothing will happen, giving us negative π to the π’.
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The second term here is π₯ squared.
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And we can find its derivative by using the power rule for differentiation.
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We multiply by the power and decrease the power by one, giving us two π₯.
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Now, we can simply substitute back in π’ is equal to π₯ squared.
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And we found that π double prime of π₯ is equal to negative two π₯π to the π₯ squared.
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And weβre ready to consider the concavity of π over the integral between zero and two.
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On this integral, since π₯ is strictly greater than zero and strictly less than two, π₯ will always have a positive value.
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And since π₯ is positive, this means that π to the π₯ squared will also be positive.
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However, negative two π₯ will be negative.
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Therefore, we have a negative number multiplied by a positive number.
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And we can say that π double prime of π₯ must be negative over our integral.
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When π double prime of π₯ is negative, this means that our function is concave down.
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So now, weβve found that π is decreasing and concave down.
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All we need to do is find if π of π₯ is greater than or less than zero.
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In order to do this, letβs consider the graph of negative π to the π₯ squared.
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This is what our graph of negative π to the π₯ squared will look like.
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We have an exponential curve, which has been flipped in the π₯-axis, due to the negative sign.
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Now, letβs consider how this graph represents π of π₯.
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Weβre mainly concerned with the values of π between the π₯-values of zero and two.
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Now, π is the integral between zero and π₯ of negative π to the π‘ squared with respect to π‘.
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Therefore, it is represented by the area between our curve and the π₯-axis between the π₯-values of zero and π₯.
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The π₯-values which weβre interested in are such that π₯ is greater than zero and less than two.
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As we can see that between the values of zero and two, this area is always below the π₯-axis.
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And since itβs below the π₯-axis, that means that our integral will evaluate to a negative number.
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And since our integral is equal to π of π₯, this means we can say that π of π₯ is less than zero.
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Therefore, we found all the information we need to know about π.
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π is decreasing.
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π is concave down.
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And π of π₯ is less than zero.
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And so our solution to this question must be option b.