Fundamental theorem of calculus

calculus Instead of doing definite integrals, we can get exact answers using the fundamental theorem of calculus.

If $f$ is continuous on [a,b] and let $F(x)={\int }_a^{x}f(t)dt$, then $F$ is a differentiable function on (a,b) and $F^{\prime }(x)=f(x)$.

Also, if $f$ is on [a,b], and F is any antiderivative of $f$, then ${\int }_a^{b}f(x)dx=F(b)-F(a)$.

… So..

I think that’s saying if you have a function f(x) and you can get it into a definite integral, then f(x) is equal to f’(x) as an indefinite integral.

Examples $F(x)={\int }_{-5}^x(t^2+sin(t))dt$. What is $F^{\prime }(x)$?

Well, that means $F^{\prime }(x)=x^{2}sin(x)$ or $F(x)=\frac{1}{3}{t}^3-cos(t)C$

---

Given ${\int }_0^{4}4x-x^{2}dx$, we use the FTOC to evaluate it.

First, take the anti-derivative.

$\frac{4}{2}{x}^2-\frac{1}{3}{x}^3+C$. We can just set $C$ to zero, because any antiderivative works.

So, we’re doing $F(b)-F(a)$.

$(2(4{)}^2-\frac{1}{3}(4{)}^3)-0=10.\bar{6}$

Also written as: ${\int }_0^4(4x-x^2)dx=(2x^2-\frac{1}{3}{x}^3){|}_0^4$.

Quiz

attempt 1

1 - WRONG

• DONE Completed: 2022-04-27 Huh. Wolfram alpha seems to agree with me?

$\begin{array}{rl}F(x)& ={\int }_1^{2}x-\frac{1}{x^3}dx\\ F^{\prime }(x)& =\frac{1}{2}{x}^2+\frac{1}{2}{x}^{-2}\\ (.5(2^2).5(2^{-2}))-(.5(1^2).5(1^{-2}))& =1.125\\ & =9/8\end{array}$

2 $\begin{array}{rl}& {\int }_0^{\frac{\pi }{4}}{\sec }^2(\theta )d\theta \\ & =\tan (x)\\ & =\tan (\frac{\pi }{4})-\tan (0)\\ & =1\end{array}$

3 - WRONG

• DONE Completed: 2022-04-27 Determine the area of the region bounded by the graphs of the line given by $y=3x+3$ and the parabola given by $y=x^2-2x+3$. Given the answer in a fraction of the lowest terms.

@axis
f(x) = x**2-2*x+3
g(x) = 3*x+3
plot f(x),g(x)

Looking at the graph, we have intersections at 0 and ~5. Validated by:

• x=0: $y=0^2-2\cdot 0+3=3$ $y=3\cdot 0+3=3$
• x=5: $y=5^2-2\cdot 5+3=25-10+3=18$ $y=3\cdot 5+3=18$

You can also determine this by solving $x^2-2x+3=3x+3$.

To segment the two, we solve: ${\int }_0^5|(x^2-2x+3)-(3x+3)|$.

$\begin{array}{rl}& =(x^2-2x+3)+-(3x+3)\\ & =x^2-2x+3-3x-3\\ & =x^2-5x+0\end{array}$

So then we solve:

$\begin{array}{rl}& ={\int }_0^5{x}^2-5x\\ & =\frac{1}{3}{x}^3-\frac{5}{2}{x}^2\\ & =\frac{1}{3}{5}^3-\frac{5}{2}{5}^2-\frac{1}{3}{0}^3-\frac{5}{2}{0}^2\\ & =41.\bar{6}-62.5-0-0\\ & =28.8\bar{3}\end{array}$

5

Suppose the acceleration of an object is given by an equation $a(t)=2+\sin (t)$

If the velocity of the object at time t=0 is $v(0)=4$, determine the velocity at time t=2. Give your answer as a decimal accurate to the nearest hundredth.

$\begin{array}{rl}a(x)& =2+\sin (x)\\ v(x)& =2x-\cos (x)+c\\ v(0)=4& =2\cdot 0-\cos (0)+c\\ & =0-1+c=4\\ c& =5\\ v(2)& =2\cdot 2-\cos (2)+5=9.42\end{array}$

attempt 2

1 - WRONG

• DONE Completed: 2022-04-27 $\begin{array}{rl}& {\int }_1^{4}x-x^{-.5}\\ & =\frac{1}{2}{x}^2-\frac{3}{2}\sqrt{x}\\ & =((.5\cdot 4^2)-(\frac{3}{2}\sqrt{4}))-((.5\cdot 1^2)-(\frac{3}{2}\cdot \sqrt{1}))\\ & =5--1\\ & =6\end{array}$

2 $\begin{array}{rl}& {\int }_0^{\frac{\pi }{2}}cos(x)dx\\ & =sin(x)\\ & =sin(\frac{\pi }{2})-sin(0)\\ & =1\end{array}$

3 - WRONG

• DONE Completed: 2022-04-27 $y=6x-3,y=x^2-x+7$

$\begin{array}{rl}6x-3=x^2-x+7& \\ 6x=x^2-x+10& \\ 0=x^2-7x+10& \\ & =\frac{7\pm \sqrt{-7^2-4\cdot 1\cdot 10}}{2\cdot 1}\\ & =2,5\\ & =(2,9),(5,27)\end{array}$

So interval is from 2-5.

$\begin{array}{rl}& {\int }_2^5|(6x-3)-(x^2-x+7)|\\ & ={\int }_2^5|x^{2}5x4|\\ & =5^2+5\cdot 5+4-2^2+5\cdot 2+4\\ & =54-18\\ & =46\end{array}$

After simplifying in step 2 (which I got wrong), we need to anti-differentiate then solve.

5 $\begin{array}{rl}a(x)& =x-\sin (x)\\ v(x)& =\frac{1}{2}{x}^2\cos (x)C\\ v(0)& =2=.5\cdot 0^2\cos (0)C\\ C& =1\\ v(x)& =.5\cdot x^2+\cos (x)+1v(3)=.5\cdot 3^2+\cos (3)+1=4.51\end{array}$