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4 years ago

Need answers to 8 and 9 fam help a girl out plz

Mathematics

1 answer:

Svetradugi [14.3K]4 years ago

7 0

For question 8, you'll have to calculate the price for three lamps and the sofa together. Take the total price of the items and multiply the percentage.
For question 9, the formula for simple interest is price(P) x rate(R) x time(T) / 100. Put in the information given:)

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a) Substitute $y=x^9$ and $dy=9x^8\,dx$ :

$\displaystyle \int x^8 \cos(x^9) \, dx = \frac19 \int 9x^8 \cos(x^9) \, dx \\\\ = \frac19 \int \cos(y) \, dy \\\\ = \frac19 \sin(y) + C \\\\ = \boxed{\frac19 \sin(x^9) + C}$

b) Integrate by parts:

$\displaystyle \int u\,dv = uv - \int v \, du$

Take $u = \ln(x)$ and $dv=\frac{dx}{x^7}$ , so that $du=\frac{dx}x$ and $v=-\frac1{6x^6}$ :

$\displaystyle \int \frac{\ln(x)}{x^7} \, dx = -\frac{\ln(x)}{6x^6} + \frac16 \int \frac{dx}{x^7} \\\\ = -\frac{\ln(x)}{6x^6} + \frac1{36x^6} + C \\\\ = \boxed{-\frac{6\ln(x) + 1}{36x^6} + C}$

c) Substitute $y=\sqrt{x+1}$ , so that $x = y^2-1$ and $dx=2y\,dy$ :

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \frac12 \int 2y e^y \, dy = \int y e^y \, dy$

Integrate by parts with $u=y$ and $dv=e^y\,dy$ , so $du=dy$ and $v=e^y$ :

$\displaystyle \int ye^y \, dy = ye^y - \int e^y \, dy = ye^y - e^y + C = (y-1)e^y + C$

Then

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \boxed{\left(\sqrt{x+1}-1\right) e^{\sqrt{x+1}} + C}$

d) Integrate by parts with $u=\sin(\pi x)$ and $dv=e^x\,dx$ , so $du=\pi\cos(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \sin(\pi x) \, e^x \, dx = \sin(\pi x) \, e^x - \pi \int \cos(\pi x) \, e^x \, dx$

By the fundamental theorem of calculus,

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = - \pi \int_0^1 \cos(\pi x) \, e^x \, dx$

Integrate by parts again, this time with $u=\cos(\pi x)$ and $dv=e^x\,dx$ , so $du=-\pi\sin(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \cos(\pi x) \, e^x \, dx = \cos(\pi x) \, e^x + \pi \int \sin(\pi x) \, e^x \, dx$

By the FTC,

$\displaystyle \int_0^1 \cos(\pi x) \, e^x \, dx = e\cos(\pi) - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx$

Then

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = -\pi \left(-e - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx\right) \\\\ \implies (1+\pi^2) \int_0^1 \sin(\pi x) \, e^x \, dx = 1 + e \\\\ \implies \int_0^1 \sin(\pi x) \, e^x \, dx = \boxed{\frac{\pi (1+e)}{1 + \pi^2}}$

e) Expand the integrand as

$\dfrac{x^2}{x+1} = \dfrac{(x^2 + 2x + 1) - (2x+1)}{x+1} = \dfrac{(x+1)^2 - 2 (x+1) + 1}{x+1} \\\\ = x - 1 + \dfrac1{x+1}$

Then by the FTC,

$\displaystyle \int_0^1 \frac{x^2}{x+1} \, dx = \int_0^1 \left(x - 1 + \frac1{x+1}\right) \, dx \\\\ = \left(\frac{x^2}2 - x + \ln|x+1|\right)\bigg|_0^1 \\\\ = \left(\frac12-1+\ln(2)\right) - (0-0+\ln(1)) = \boxed{\ln(2) - \frac12}$

f) Substitute $e^{7x} = \tan(y)$ , so $7e^{7x} \, dx = \sec^2(y) \, dy$ :

$\displaystyle \int \frac{e^{7x}}{e^{14x} + 1} \, dx = \frac17 \int \frac{\sec^2(y)}{\tan^2(y) + 1} \, dy \\\\ = \frac17 \int \frac{\sec^2(y)}{\sec^2(y)} \, dy \\\\ = \frac17 \int dy \\\\ = \frac y7 + C \\\\ = \boxed{\frac17 \tan^{-1}\left(e^{7x}\right) + C}$