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secondary 4 | A Maths
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morning everyone! I need help with both parts of this question, Thank you.
Date Posted: 4 years ago
Views: 329
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Good afternoon Sonia! Here are my solutions to the question. I have included the diagram as well for your reference, which I have cross-checked with the Desmos graphing calculator online to ensure the shape of my graphs are correct (except that the straight line actually passes through the maximum point at x = 1, sorry!)
Let me know if you do not understand my workings and I will do my best to explain them in more detail!
Let me know if you do not understand my workings and I will do my best to explain them in more detail!
Date Posted:
4 years ago
The techniques to solve this question are as follows.
1. We must find at least the x-coordinates of the points of intersection (unless question asks for the full coordinates) if we plan to integrate the expression with respect to x (which we usually do), because we need the values of x as the integration limits.
2. We must next ensure that the graphs are above the x-axis, because integrals become negative for parts of graphs below the x-axis and this interferes with the area above the x-axis, so care needs to be taken here.
3. In shaded regions bounded by only one curve, the area under the graph is simply the definite integral of the equation of the curve within the specified integral limits.
4. In shaded regions such as this question which are bounded by two curves, we need to find out which curve is residing above which curve (assuming both are above the x-axis), because this is important in obtaining the area value. Fortunately, if we put the wrong way round, the area will be the negative of the correct area, of which the magnitude will of course be the same. If you are scared, you can apply modulus as well to remove any potential negative signs arising from the computation.
5. And there you have it, the area of the shaded regions.
1. We must find at least the x-coordinates of the points of intersection (unless question asks for the full coordinates) if we plan to integrate the expression with respect to x (which we usually do), because we need the values of x as the integration limits.
2. We must next ensure that the graphs are above the x-axis, because integrals become negative for parts of graphs below the x-axis and this interferes with the area above the x-axis, so care needs to be taken here.
3. In shaded regions bounded by only one curve, the area under the graph is simply the definite integral of the equation of the curve within the specified integral limits.
4. In shaded regions such as this question which are bounded by two curves, we need to find out which curve is residing above which curve (assuming both are above the x-axis), because this is important in obtaining the area value. Fortunately, if we put the wrong way round, the area will be the negative of the correct area, of which the magnitude will of course be the same. If you are scared, you can apply modulus as well to remove any potential negative signs arising from the computation.
5. And there you have it, the area of the shaded regions.
Thank you so much Eric!!! Your explanations and detail in your working is amazing, Thank you for the effort and time :)
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