bigmansouf
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My question is
FInd the areas enclosed by the following curves and straight lines
a)  y=x^{3}-1 , the axes, y=26

Attachment 473187473189473200

MY answer.
Find area A
 \int_{0}^{26}(y+1)^{\frac{1}{3}} dy
 [\frac{(y+1)^{\frac{4}{3}}}{\frac{4}{3}}]_{0}^{26}
 [\(y+1)^{\frac{4}{3}}) \times \frac{3}{4}]_{0}^{26}
 [\frac{3(y+1)^{\frac{4}{3}})}{4}]_{0}^{26}
 (\frac{3(26+1)^{\frac{4}{3}}}{4}) - (\frac{3(0+1)^\frac{4}{3})}{4}) = \frac{243}{4} -\frac{3}{4} = 60

Area A= 60

Find Area B

 \int_{0}^{26}(y+1)^{\frac{1}{3}} dy
 [\frac{(y+1)^{\frac{4}{3}}}{\frac{4}{3}}]_{-1}^{0}
 [\(y+1)^{\frac{4}{3}}) \times \frac{3}{4}]_{-1}^{0}
 [\frac{3(y+1)^{\frac{4}{3}})}{4}]_{-1}^{0}
 (\frac{3(0+1)^\frac{4}{3})}{4}) - (\frac{3(-1+1)^\frac{4}{3})}{4})

Area B=  \frac{3}{4}

Therefore Area required is 59.25= 60-3/4

However I got the wrong answer the book says the answer is 60.

I dont understand why because i thought the question asks me to find the area that is shaded pink in the picture (attached) (encaptioned my thoughts)
but is seems that the area the book wanted me to find the area coloured green which is the pitcure (encaptioned book's thoughts). I am confused as i thought I had to calculate the axes as as the straight line given. Therefore, I thought I had to calculate the area marked (area G) as it sits on the y-axis
Can someone please explain what I did wrong?

Thank you

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B_9710
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(Original post by bigmansouf)
My question is
FInd the areas enclosed by the following curves and straight lines
a)  y=x^{3}-1 , the axes, y=26

Attachment 473187473189473200

MY answer.
Find area A
 \int_{0}^{26}(y+1)^{\frac{1}{3}} dy
 [\frac{(y+1)^{\frac{4}{3}}}{\frac{4}{3}}]_{0}^{26}
 [\(y+1)^{\frac{4}{3}}) \times \frac{3}{4}]_{0}^{26}
 [\frac{3(y+1)^{\frac{4}{3}})}{4}]_{0}^{26}
 (\frac{3(26+1)^{\frac{4}{3}}}{4}) - (\frac{3(0+1)^\frac{4}{3})}{4}) = \frac{243}{4} -\frac{3}{4} = 60

Area A= 60

Find Area B

 \int_{0}^{26}(y+1)^{\frac{1}{3}} dy
 [\frac{(y+1)^{\frac{4}{3}}}{\frac{4}{3}}]_{-1}^{0}
 [\(y+1)^{\frac{4}{3}}) \times \frac{3}{4}]_{-1}^{0}
 [\frac{3(y+1)^{\frac{4}{3}})}{4}]_{-1}^{0}
 (\frac{3(0+1)^\frac{4}{3})}{4}) - (\frac{3(-1+1)^\frac{4}{3})}{4})

Area B=  \frac{3}{4}

Therefore Area required is 59.25= 60-3/4

However I got the wrong answer the book says the answer is 60.

I dont understand why because i thought the question asks me to find the area that is shaded pink in the picture (attached) (encaptioned my thoughts)
but is seems that the area the book wanted me to find the area coloured green which is the pitcure (encaptioned book's thoughts). I am confused as i thought I had to calculate the axes as as the straight line given. Therefore, I thought I had to calculate the area marked (area G) as it sits on the y-axis
Can someone please explain what I did wrong?

Thank you

Name:  q6.png
Views: 44
Size:  25.7 KBAttachment 473187473189
The question is asking you to find the area in green. The area you are working out is the area bounded by the line y=26 , the curve and the coordinate axes. This means the area in green. If it never said the area bounded by the coordinate axes and just said find the area bounded by the curve and the line y=26 and the y axis then the area you would be trying to find would be the area in pink. Because it explicitly says the coordinate axes - this means the x axis and the y axis, not just the y axis.
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bigmansouf
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(Original post by B_9710)
The question is asking you to find the area in green. The area you are working out is the area bounded by the line y=26 , the curve and the coordinate axes. This means the area in green. If it never said the area bounded by the coordinate axes and just said find the area bounded by the curve and the line y=26 and the y axis then the area you would be trying to find would be the area in pink. Because it explicitly says the coordinate axes - this means the x axis and the y axis, not just the y axis.
Thanks you very much for the quick reply
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