# Calculus 1 basic level questions ## Question 1 #### $0\int^5$ $x/(1+x^2)^3dx$ on the integral from 0 to 2 ## solution: #### Let u= $1+x^2$ #### Let du= $2xdx$ #### $1/2du=xdx$ ### $1\int^5$ $1/u^3*1/2$ #### $1/2* 1\int^5$$u^-3 du$ #### $1/2-1/2u^-2$ on the integral from 1 to 5 #### $1/2-1/2(1+x^2)$ on the integral from 1 to 5 ### This is equal to 6/25 ##### Therefore by using u substitution of the above function from those intervals, the derivative is 6/25. <iframe src="https://www.desmos.com/calculator/7yrmkbgrcm?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe> # Question 2 ## We are going to state the two forms of the fundamental theorem of calculus. ##### Form 1: ###### If function G which is an antiderivative of f on [a,b] (G'(x)=f(x)) #### Then: $$a\int^b f(x)dx=G(b)-G(a)=G(x)\mid^ba$$ ## This forms main use is for computing integrals #### Example: #### $$0\int^3 \mid x^2-1/ dx$$ #### $$0\int^1 \mid x^2-1/dx+1\int^3 \mid x^2-1/dx$$ #### $$0\int^1 1-x^2$$ ##### $$x-x^3/3$$ on the integral from 0 to 1 = 1-1/3 =2/3 #### $$1\int^3x^2-1 dx =x^3/x-x$$ on the interval form 1 to 3. This equals $$ (3^3/3-3)-(1^3/3-1)=6$$ ### $$ 2/3+6=22/3$$ which is the final answer. #### Form 2: ### If f(x) is continuous on [a,b], you can define a new function: #### $$A(x)=a\int^xf(t)dt$$, then $$A'(x)=f'(x)$$Aslong as x stays between [a,b]. The purpose of this form is to use it when you need to find the derivative of the integral. ### Example of form 2: #### $$G(x)=-2x\int^xsec^2tdt$$ #### $$-2x\int^0sec^2tdt+0\int^xsec^2tdt$$ ### Simplifying this using A:A(x)=$$0\int^xsec^2tdt$$A'(X)=$$sec^2t$$ #### $$-A(-2x)+A(x) then G'(x)=-A(-2x)(-2)+A'(x)$$ ### Final answer: $$2*sec^2 (-2x)+sec^2x$$ ## Question 3-Define two forms of the limit definition of a derivative. #### Form 1: ## $$f'(x)$$=lim as h goes to 0 $$f(x+h)-f(x)/h$$ ### solve for $$x^2 +1$$ as h goes to 0 $$((x+h^2)+1)-(x^2+1)/h$$ #### $$(x^2 +2xh+h^2 -x^2+1)/h$$ #### $$(2xh+h^2)/h ### $$h(2x+h)/h$$ ### the limit as h goes to 0 is $$2x+h$$ setting h to 0 makes the final answer of f'(x)=2x ## Form 2: ### Definition: limit as x goes to a $$f(x)-f(a)/x-a$$ ### using $$x^2+1$$ ## $$(x^2+1)-(a^2+1)/x-a$$ ### $$d/dx(x^2 +1-a^2 +1)/x-a$$ #### $$(2x+-d/dxa^2 +1)/d/dx(x-a)$$ ### $$2x-0/1$$ ## Using form two as the limit of x goes to a, the final answer (derivative) =$$2x$$