Intergrals (sums and differences in area) (1 Viewer)

V_L · Mar 26, 2017

Hi,

Im sorry i know this is a shit load of questions but I've been struggling all afternoon with them so any help would be greatly appreciated!!

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9. Find the area enclosed between the curve y = x 3 , the x -axis and the line y = -3 x + 4

10. Find the area enclosed by the curves y = ( x - 2)^2 and y = ( x - 4)^2 .

11. Find the area enclosed between the curves y = x^2 and y = x^3 .

12. Find the area enclosed by the curves y = x^2 and x = y62 .

13. Find the area bounded by the curve y = x^2 + 2 x - 8 and the line y = 2 x + 1.

14. Find the area bounded by the curves y = 1 - x^2 and y = x^2 - 1.

15. Find the exact area enclosed between the curve y = √(4 − x^2) and the line x - y + 2 = 0.

Answers:

9. 0.42 units^2
10. ⅔ units^2
11. 1/12 units^2
12. ⅓ units^2
13. 36 units^2
14. 2⅔ units^2
15. (π - 2) units^2

Drongoski · Mar 26, 2017

Q9) The 2 curves y = x^3 and y = -3x + 4 meet at x = 1. The reqd area is made up of 2 parts: the area under the 1st curve from x = 0 to x = 1; the 2nd part is the area under the straight line from x = 1 to x = 4/3. Therefore reqd area:

$= \int ^1 _ 0 x^3 dx + \int ^{\frac {4}{3}} _ 1 (-3x + 4)dx \\ \\ = \frac {5}{12} = 0.416666 . . .$

Drongoski · Mar 26, 2017

Q10) The 2 curves intersect at x = 3 and touch the x-axis at x= 2 and x = 4 resp. The reqd area is made up of 2 parts (by symmetry, each of same area)
$= \int ^3 _2 (x-2)^2 dx + \int ^4 _3 (x-4)^2 dx = \left [\frac {(x-2)^3}{3} \right ] ^3 _ 2 + \left [\frac {(x-4)^3}{3} \right ]^4 _3 = \frac {2}{3}$

1729 · Mar 26, 2017

V_L said:
Hi,

Im sorry i know this is a shit load of questions but I've been struggling all afternoon with them so any help would be greatly appreciated!! )

9. Find the area enclosed between the curve y = x 3 , the x -axis and the line y = -3 x + 4

10. Find the area enclosed by the curves y = ( x - 2) 2 and y = ( x - 4) 2 .

11. Find the area enclosed between the curves y = x 2 and y = x 3 .

12. Find the area enclosed by the curves y = x 2 and x = y 2 .

13. Find the area bounded by the curve y = x 2 + 2 x - 8 and the line y = 2 x + 1.

14. Find the area bounded by the curves y = 1 - x 2 and y = x 2 - 1.

15. Find the exact area enclosed between the curve y = 4 − x2 and the line x - y + 2 = 0.

Answers:

9. 0.42 units^2
10. ⅔ units^2
11. 1/12 units^2
12. ⅓ units^2
13. 36 units^2
14. 2⅔ units^2
15. (π - 2) units^2

$\noindent \textbf{Q15.} $A = \int_{-2}^1(4-x^2-(x+2))\mathrm{d}x = \int_{-2}^1(4-x^2-x-2)\mathrm{d}x = \int_{-2}^1(2-x^2-x)\mathrm{d}x = \left[2x - \frac{1}{3}x^3 - \frac{1}{2}x^2\right]^1_{-2} = 2 - \frac{1}{3} - \frac{1}{2} - 2(-2) + \frac{1}{3}(-2)^3 + \frac{1}{2}(-2)^2 = 4\frac{1}{2}$ units$^2$

$\noindent \textbf{Q14.} By symmetry, $A = 4\int_0^1(1-x^2)\mathrm{d}x = 4\left[x - \frac{1}{3}x^3\right]^1_0 = 4\left(1 - \frac{1}{3} - 0\right) = 2\frac{2}{3}$ units$^2$

$\noindent \textbf{Q13.} $A = \int_{-3}^3(2x+1 - (x^2+2x-8))\mathrm{d}x = \int_{-3}^3(2x+1-x^2-2x+8)\mathrm{d}x = \int_{-3}^3(9 - x^2)\mathrm{d}x = \left[9x - \frac{1}{3}x^3\right]^3_{-3} = 9(3) -\frac{1}{3}(3)^3 - 9(-3) + \frac{1}{3}(-3)^3 = 36$ units$^2$

$$\noindent \textbf{Q12.} $A = \int_{0}^1(\sqrt{x} - x^2)\mathrm{d}x = \left[\frac{2x^{\frac{3}{2}}}{3} - \frac{1}{3}x^3\right]^1_0 = \frac{2(1)^{\frac{3}{2}}}{3} - \frac{1}{3}(1)^3 = \frac{1}{3}$ units$^2$

$$\noindent \textbf{Q11.} $A = \int_{0}^1(x^2-x^3)\mathrm{d}x=\left[\frac{1}{3}x^3 - \frac{1}{4}x^4\right]^1_0 = \frac{1}{3}(1)^3 - \frac{1}{4}(1)^4 = \frac{1}{12}$ units$^2$

$$\noindent \textbf{Q10.} By symmetry, $A = 2\int_2^3(x-2)^2\mathrm{d}x$. This is equivalent, by translation, to $A = 2\int_0^1x^2\mathrm{d}x = 2\left[\frac{1}{3}x^3\right]^1_0=2\left(\frac{1}{3}(1)\right) = \frac{2}{3}$ units$^2$

$$\noindent \textbf{Q9.} $A = \int_0^1x^3\mathrm{d}x + \frac{1}{2}bh = \left[\frac{1}{4}x^4\right]^1_0 + \frac{1}{2}\left(\frac{1}{3}\right)(1) = \frac{1}{4}(1) + \frac{1}{6} = \frac{5}{12}$ units$^2$

V_L · Mar 27, 2017

Hi, Thank you so so much!! Could you please explain how Q10 works please? Also i made a mistake on question 15, one of the equations was meant to be y = the square root of (4-x^2) so thats why the answers don't match

1729 · Mar 27, 2017

V_L said:
Hi, Thank you so so much!! Could you please explain how Q10 works please? Also i made a mistake on question 15, one of the equations was meant to be y = the square root of (4-x^2) so thats why the answers don't match

$$\noindent For Q10, the area required is given by $A = \int_2^3(x-2)^2\mathrm{d}x + \int_3^4(x-4)^2\mathrm{d}x.$ However, by symmetry these two areas are equal and thus we can say $A = 2\int_2^3(x-2)^2\mathrm{d}x$. \\\\ Now, this can be evaluated as it is, but, to avoid more substitution, notice that if the graph of $y=(x-2)^2$ is shifted onto the origin, we have $y=x^2$. In doing so, we shift the boundaries $x=2, x=3$ in the same manner to obtain $x=0, x=1$. \\\\ Thus, we can say $A = 2 \int_0^1x^2\mathrm{d}x$ (a more elegant expression) which is simply equivalent in area to $A = 2\int_2^3(x-2)^2\mathrm{d}x$, by translation.$

1729 · Mar 27, 2017

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