redpillowsparkle
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Integral calculus (1 Viewer)
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redpillowsparkle
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2nd - question
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So the integral of a function calculates the signed area. area above the x-axis is positive and area below x-axis is negative You want find the value of k so that area above the x-axis equals the area below the x-axis
redpillowsparkle
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Last question
Most of the question you seem to be asking seem to standard math questions. Terry Lee is good textbook to use to study
I haven't look at it for 2u. I would presume it would be good.
Solutions.
8. Solve the equation from the areas of the 2 trapezia(10)/2=(6+4)(k-4)/2\Rightarrow k=11\therefore )
D.
You can also work it out using the integrals of the lines, but using the areas of the trapezia is easier.
10.^2-(\frac{x}{4})^2)dx=\pi\int_0^{64}(4x-\frac{x^2}{16})dx\therefore )
D.
13ai.dx=\frac{12x^6}{6}+\frac{7x^4}{4}-9x+c=2x^6+\frac{7x^4}{4}-9x+c)
13aii.
13b. dx=\int_1^2(6x^2-x)dx=\left\[2x^3-\frac{x^2}{2}\right\]_1^2)
=12.5)
13e.^2dy=\pi\int_0^1(y^2-2y^3+y^4)dy=\pi\left\[\frac{y^3}{3}-\frac{y^4}{2}+\frac{y^5}{5}\right\]_0^1)
=\frac{\pi}{30})
14a.dx+\int_3^4(-x^2+4x)dx=\left\[\frac{-x^3}{3}+4x^2-12x\right\]_2^3+\left\[\frac{-x^3}{3}+2x^2\right\]_3^4)
-\frac{64}{3}+32-(-9+18)=\frac{10}{3})
14bi.=\frac{(5x^2+1)(2x)-(x^2-3)(10x)}{(5x^2+1)^2}=\frac{32x}{(5x^2+1)^2})
14bii.^2}dx=\frac{x^2-3}{5x^2+1}+c)
^2}dx=\frac{1}{2}\cdot\frac{x^2-3}{5x^2+1}+c=\frac{x^2-3}{10x^2+2}+c)
These concur with answers I got before with wolframalpha.
8. Solve the equation from the areas of the 2 trapezia
You can also work it out using the integrals of the lines, but using the areas of the trapezia is easier.
10.
13ai.
13aii.
13b.
13e.
14a.
14bi.
14bii.
These concur with answers I got before with wolframalpha.
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