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Find the area of the region above the $x$-axis bounded by the line $y=4x$ and the curve $y=x^3$

Attempt:

intersect when:- $x^3 - 4x = 0$

$x ( x² - 4 ) = 0 $

$x = 0 , x = \pm 2$

Area is given by :-

$$2 ∫_0^2 4x - x^3\; dx = 2 [ 2x^2 - \frac13x^3 ]_0^2$$

$$2 [ 8 - 8/3 ] = 32/3\;\;\text{units²}$$

I want to understand this topic well so I'm solving different questions from textbooks under it. The answer in the textbook for this is 6.75 units². I can't figure where I'm wrong.

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  • $\begingroup$ If you show your attempts it will be easier to help you find where you are wrong. $\endgroup$ – uniquesolution Oct 10 '15 at 11:14
  • $\begingroup$ I showed my attempt. Why the down vote? Look well. $\endgroup$ – user274246 Oct 10 '15 at 11:19
  • $\begingroup$ Oh, I missed your attempt, probably because your formatting is not very good. Please forgive me for not trying to read it. $\endgroup$ – uniquesolution Oct 10 '15 at 11:20
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    $\begingroup$ Alright. Please can you upvote me back? $\endgroup$ – user274246 Oct 10 '15 at 11:26
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    $\begingroup$ I use an android phone. So I can't writel wirh latex. I'm sorry $\endgroup$ – user274246 Oct 10 '15 at 11:39
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enter image description here

So enclosed area $\displaystyle = \int_{0}^{2}\left[4x-x^3\right]dx = $

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  • $\begingroup$ Thanks a lot for answering but It's still not giving me the answer. Can the mistake be from the textbook? $\endgroup$ – user274246 Oct 10 '15 at 13:08
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your questions asks for the area above the x-axis, that means you just ignore the area under the x-axis that's bound by the line and the curve.

sorry for the kind of bad graph

and it's found by :

$Area = \int_{0}^2 [4x - x^3]dx = [2x^2 - \frac{1}{4}x^4]^2_0 = 2(2)^2 - \frac{1}{4} (2)^4 - (2(0)^2-\frac{1}{4}(0)^4) = 4$ $units^2$

the answer in your book is probably wrong, though if you could check with someone that has/knows that book as well that'd be better.

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The real problem is that $\displaystyle \int x^3 = \frac{x^4}{4}$, not $\displaystyle \frac{x^3}{3}$.

$$\int_0^2 4x-x^3$$ $$= \left[4 \cdot \frac{x^2}{2} - \frac{x^\color{red}{4}}{\color{red}{4}} \right]_0^2$$ $$= \left[ 2x^2 - \frac{x^4}{4} \right]_0^2$$ $$= \left(2\cdot 2^2 - \frac{2^4}{4} \right) - \left(2\cdot 0^2 - \frac{0^4}{4} \right)$$ $$ = (8 - 4) - (0)$$ $$ = 4$$

This agrees with the answer given by Wolfram Alpha, so your textbook is probably wrong. This shows that even textbooks make mistakes!

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