009B Sample Midterm 2, Problem 2

This problem has three parts:

a) State the Fundamental Theorem of Calculus.

b) Compute   ${\frac {d}{dx}}\int _{0}^{\cos(x)}\sin(t)~dt$.

c) Evaluate $\int _{0}^{\pi /4}\sec ^{2}x~dx$.

Foundations:
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 $G(x)=\int _{0}^{x}\sin(t)~dt$.
Then, $F(x)=G(g(x))$.

Step 2:
If we take the derivative of both sides of the last equation, we get $F'(x)=G'(g(x))g'(x)$ by the Chain Rule.

Step 3:
Now, $g'(x)=-\sin(x)$ and $G'(x)=\sin(x)$ by the Fundamental Theorem of Calculus, Part 1.
Since $G'(g(x))=\sin(g(x))=\sin(\cos(x))$, we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\frac {\pi }{4}}\sec ^{2}x~dx=\tan(x){\biggr |}_{0}^{\pi /4}}

Step 2:
So, we get
$\int _{0}^{\frac {\pi }{4}}\sec ^{2}x~dx=\tan {\bigg (}{\frac {\pi }{4}}{\bigg )}-\tan(0)=1$.

Final Answer:
(a)
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 $F'(x)=f(x)$.
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)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}\int _{0}^{\cos(x)}\sin(t)~dt\,=\,\sin(\cos(x))\cdot (-\sin(x))} .
(c) 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\,=\,1} .

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