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 $F(x)=\frac{1}{3}{t}^3-cos(t)+C$

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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

2

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

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

2

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

5

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