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The figure shows the graph of π of π₯ equals a quarter π₯ minus two squared π₯ plus one.
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Calculate the area of the shaded region, giving your answer as a fraction.
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To calculate the area between a curve and the π₯-axis, we need to integrate that function with respect to π₯.
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In this case, weβll be integrating the function a quarter π₯ minus two squared π₯ plus one with respect to π₯.
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And thatβs between the limits two and negative one, since those are the π₯-values which represent the upper and lower limits of the area required.
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We do, however, need to be a little bit careful since integrating a function whose area lies underneath the π₯-axis will give a negative value.
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In this case so, weβre purely interested in the area above the π₯-axis.
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So, weβre fine just to integrate here.
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And the easiest way to integrate this function, in my opinion, is to distribute the parentheses.
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We could use a substitution of π’ equals π₯ minus two, but itβs simple enough to just read these parentheses.
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So, letβs go ahead and do that.
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We begin by distributing π₯ minus two all squared.
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By remembering that we need to write this as the product of two brackets, π₯ minus two and π₯ minus two.
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π₯ multiplied by π₯ is π₯ squared.
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We multiply the outer terms.
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We get negative two π₯.
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Then we multiply the inner terms, and we get negative two π₯ again.
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Then multiplying the last terms, we get positive four.
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So, we can see π₯ minus two all squared is equal to π₯ squared minus four π₯ plus four.
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Weβre then going to multiply this by π₯ plus one, being really careful to multiply each term in the first parenthesis by each term in the second.
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One way to do this is to use the grid method.
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π₯ multiplied by π₯ squared is π₯ cubed.
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And π₯ squared multiplied by one is π₯ squared.
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π₯ multiplied by negative four π₯ is negative four π₯ squared.
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And negative four π₯ multiplied by one is negative four π₯.
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π₯ multiplied by four is four π₯.
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And four multiplied by one is four.
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If we collect like terms, we can see that π₯ minus two all squared multiplied by π₯ plus one is π₯ cubed minus three π₯ squared plus four.
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So, we need to actually integrate a quarter π₯ cubed minus three π₯ squared plus four with respect to π₯ between those limits two and negative one we said earlier.
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In fact, whenever we have a function in π₯ is being multiplied by some constant, it can be simpler to bring this constant outside of the integration sign.
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This is allowed.
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We will simply end up multiplying by a quarter at the end, rather than at the beginning of our calculations.
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Letβs integrate π₯ cubed minus three π₯ squared plus four.
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To integrate π₯ cubed, we add one to the exponent.
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That gives us π₯ to the power of four.
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We then divide through by the value of that new exponent.
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So, π₯ cubed becomes π₯ to the power of four divided by four.
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Negative three π₯ squared becomes negative three π₯ cubed divided by three, which is simply negative π₯ cubed.
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And four integrates to become four π₯.
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Remember we do not need the constant of integration when integrating between two limits.
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Weβll just need to evaluate between the limits two and negative one.
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So, we substitute two and negative one into our expression and find the difference between them.
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Substituting two in, we get two to the power of four divided by four minus two cubed plus four multiplied by two, which is four.
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And the second part is negative one to the power of four divided by four minus negative one cubed plus four multiplied by negative one.
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And that simplifies to negative 11 over four.
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And if we type all of this into our calculator, we get an answer of 27 over 16.
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And so, we can see that the shaded area is 27 over 16 square units.