Answer to Question #198718 in Calculus for desmond

Question

3. (a) Use the Riemann sum to calculate the area under the curve y = −x 2 + 3x + 2 between x = −1 and x = 2. 

b.(b) Let R = \lim _{n\to \infty \:}\left(\sum _{j=1}^{n-1}\:\left(3+\frac{4j}{n}\right)^2.\frac{4}{n}\right), write R as a definite integral and hence evaluate R

c.) If x^y\:=\:e^{x-y}, prove that \frac{d}{dx}=\frac{ln\:x}{\left(1+lnx\right)^2}

Answer 1

Part a.

$∫abf(x)dx≈Δx(f(x0)+f(x1)+f(x2)+⋯+f(xn−2)+f(xn−1))$

$where Δx=\frac{b−a}{n}$

We have that $a=−1, b=2, n=4.$

$Therefore, Δx=\frac{2−(−1)}{4}=3/4$

Divide the interval $[−1,2]$  into n=4 subintervals of the length Δx=3/4 with the following endpoints $a=−1 , −1/4, 1/2, 5/4, 2=b$

Now, just evaluate the function at the left endpoints of the subintervals.

$f(x0)=f(−1)=0$

$f(x1)=f(−1/4)=21/16=1.3125$

$f(x2)=f(1/2)=15/4=3.75$

$f(x3)=f(5/4)=117/16=7.3125$

Finally, just sum up the above values and multiply by $Δx=3/4$

$3/4(0+1.3125+3.75+7.3125)=9.28125$

Part b

$\lim _{n\to \infty \:}\left(\sum _{i=1}^n\left(3+\frac{4n}{n}\right)^2\frac{4}{n}\right)$

$\mathrm{The\:definite\:integral\:is\:defined\:as:\:}\lim _{n\to \infty }\left(\sum _{i=1}^n\left(f\left(c_i\right)\Delta \:x\right)\right)=\int _a^bf\left(x\right)dx$

$\mathrm{Where\:}Δx=\frac{b-a}{n},\:\mathrm{and}\:c_i=a+Δx⋅i$

$=\int _0^1196dx$

$=196$

Part c

$=e^{x-y}\frac{d}{dx}\left(x-y\right)$

$=1-\frac{d}{dx}\left(y\right)$

$=e^{x-y}\left(1-\frac{d}{dx}\left(y\right)\right)$

$\left(\frac{\ln \left(x\right)}{\left(1+\ln \left(x\right)\right)^2}\right)$