007B Sample Midterm 2, Problem 1 Detailed Solution

This problem has three parts:

(a) State both parts of the fundamental theorem of calculus.

(b) Compute   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.}

(c) Evaluate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\pi /4}\sec ^{2}x~dx.}

Background Information:
1. What does Part 1 of the Fundamental Theorem of Calculus say about  ${\frac {d}{dx}}\int _{0}^{x}\sin(t)~dt?$
Part 1 of the Fundamental Theorem of Calculus says that
${\frac {d}{dx}}\int _{0}^{x}\sin(t)~dt=\sin(x).$
2. What does Part 2 of the Fundamental Theorem of Calculus say about  $\int _{a}^{b}\sec ^{2}x~dx$  where  $a,b$  are constants?
Part 2 of the Fundamental Theorem of Calculus says that
$\int _{a}^{b}\sec ^{2}x~dx=F(b)-F(a)$  where  $F$  is any antiderivative of  $\sec ^{2}x.$

Solution:

(a)

Step 1:
The Fundamental Theorem of Calculus has two parts.
The Fundamental Theorem of Calculus, Part 1
Let  $f$  be continuous on  $[a,b]$  and let  $F(x)=\int _{a}^{x}f(t)~dt.$
Then,  $F$  is a differentiable function on  $(a,b)$  and  $F'(x)=f(x).$

Step 2:
The Fundamental Theorem of Calculus, Part 2
Let  $f$  be continuous on  $[a,b]$  and let  $F$  be any antiderivative of  $f.$
Then,
$\int _{a}^{b}f(x)~dx=F(b)-F(a).$

(b)

Step 1:
Let  $F(x)=\int _{0}^{\cos(x)}\sin(t)~dt.$
The problem is asking us to find  $F'(x).$
Let  $g(x)=\cos(x)$  and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(x)=\int_0^x \sin(t)~dt.}
Then,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(x)=G(g(x)).}

Step 2:
If we take the derivative of both sides of the last equation,
we get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F'(x)=G'(g(x))g'(x)}
by the Chain Rule.

Step 3:
Now,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)=-\sin(x)}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G'(x)=\sin(x)}
by the Fundamental Theorem of Calculus, Part 1.
Since
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G'(g(x))=\sin(g(x))=\sin(\cos(x)),}
we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x)).}

(c)

Step 1:
Using the Fundamental Theorem of Calculus, Part 2, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}.}

Step 2:
So, we get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1.}

Final Answer:
(a)     See solution above.
(b)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(\cos(x))\cdot(-\sin(x))}
(c)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1}

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