AP Calculus BC - Fundamental Theorem of Calculus with Definite Integrals

Evaluate $\int_{0}^{6} (x-3)^2 dx$

$\frac{1}{3}$

Explanation

$\int (x-3)^2 dx = \frac{1}{3} (x-3)^3 + C$

Use the fundamental theorem of calculus to evaluate:

$\frac{1}{3} (6-3)^3 - [ \frac{1}{3}(0-3)^3] = \frac{1}{3} (3)^3 - \frac{1}{3} (-3)^3$

$9 + \frac{1}{3}(27) = 9+9 = 18$

Evaluate :

$f(x)=\int_{0}^{x} \sin\left(\sqrt{t^2+\frac{\pi^2}{4}} \right )dt$

$\frac{\pi}{2}$

$-\frac{\pi}{2}$

Explanation

By the Fundamental Theorem of Calculus, we have that ${f'(x) = \frac{d}{dx} \int_0^x \sin \sqrt{t^2 + \frac{\pi ^2}{4}} dt = \sin \sqrt{x^2 + \frac{\pi ^2}{4}}$ . Thus, $f'(0) = \sin \sqrt{0^2 + \frac{\pi ^2}{4}} =1$ .

$\int _{\frac{\pi}{3}} ^ {\pi } \sin (x/2) dx$

$\sqrt3$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{4}$

$2 \sqrt3$

Explanation

$\int \sin (x/2) dx = -2 \cos (x/2 )$

Use the Fundamental Theorem of Calculus and evaluate the integral at both endpoints:

$-2 \cos (\pi/2 ) - -2 \cos (\frac{\pi}{3} / 2 ) = -2 \cos (\frac{\pi}{2}) + 2 \cos (\frac{\pi}{6})$

$= -2(0) + 2(\frac{\sqrt3}{2} )= \sqrt3$

Suppose we have the function

$\small g(x)=\int_{0}^{\cos x}(1+\sin t)dt$

What is the derivative, $\small g'(x)$ ?

$\small \small -\sin x\left[1+\sin\left(\cos(x) \right )\right]$

$\small \small \small \small \sin x\left[1+\sin\left(\cos x \right )\right]$

$\small \small \small 1+\sin\left(\cos(x) \right )$

$\small \small \small 1-\sin\left(\cos(x) \right )$

Explanation

We can view the function $\small g(x)$ as a function of $\small \cos x$ , as so

$\small g(x)=F(\cos x)$

where $\small F(x)=\int_0^x (1+\sin t)dt$ .

We can find the derivative of $\small g(x)$ using the chain rule:

$\small g'(x)=F'(\cos x)\cdot (\cos x)'=F'(\cos x)\cdot (-\sin x)$

where $\small F'(z)$ can be found using the fundamental theorem of calculus:

$\small \small F'(x)=\frac{d}{dz}\int_0^z (1+\sin t)dt=1+\sin z$

So we get

$\small \small \small \small g'(x)=-\sin x\cdot F'(\cos x)=-\sin x\cdot[1+\sin(\cos x)]$

Evaluate when $f(x)=\int_{0}^{x}(2t^{2}+5t-4)dt$ .

Explanation

Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_{0}^{x}(2t^{2}+5t-4)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(2t^{2}+5t-4)dt=2x^{2}+5x-4$ .

Therefore $f'(4)=2(4)^{2}+5(4)-4=32+20-4=52-4=48$ .

$\int_{-1}^{3} \left (4x^2 - x \right )dx$

$33.\overline{3}$

$0.\overline{3}$

$32.\overline{3}$

$31.\overline{6}$

Explanation

$\int \left ( 4x^2 -x\right ) dx = \frac{4}{3} x^3 - \frac{1}{2} x + C$

Use the Fundamental Theorem of Calculus and evaluate the integral at both endpoints:

$\frac{4}{3} (3)^3 - \frac{1}{2} (3)^2 - [ \frac{4}{3} (-1)^3 - \frac{1}{2} (-1)^2]$

$= 36 - 4.5 - [ -\frac{4}{3} - \frac{1}{2} ] = 36 - 4.5 + 1. \overline{3} + .5 = 33. \overline{3}$

Given

$f(x)=\int_{0}^{x}(\frac{2}{t^{2}+4t-5})dt$ , what is ?

$\frac{2}{7}$

$-\frac{2}{7}$

$-\frac{7}{2}$

$\frac{7}{2}$

None of the above.

Explanation

By the Fundamental Theorem of Calculus, for all functions that are continuously defined on the interval with in and for all functions defined by by $F(x)=\int_{a}^{x}f(t)dt$ , we know that .

Given

$f(x)=\int_{0}^{x}\left(\frac{2}{t^{2}+4t-5}\right)dt$ , then

$f'(x)=\frac{d}{dx}\int_{0}^{x}\left(\frac{2}{t^{2}+4t-5}\right)dt=\frac{2}{x^{2}+4x-5}$ .

Therefore,

$f'(2)=\frac{2}{2^{2}+4(2)-5}=\frac{2}{4+8-5}=\frac{2}{4+3}=\frac{2}{7}$ .

Given

$f(x)=\int_{0}^{x}\left(\frac{1}{t^{2}}-2t+3\right)dt$ , what is ?

$-\frac{323}{36}$

$\frac{323}{36}$

$-\frac{36}{323}$

$\frac{36}{323}$

None of the above.

Explanation

By the Fundamental Theorem of Calculus, for all functions that are continuously defined on the interval with in and for all functions defined by by $F(x)=\int_{a}^{x}f(t)dt$ , we know that .

Thus, for

$f(x)=\int_{0}^{x}\left(\frac{1}{t^{2}}-2t+3\right)dt$ ,

$f'(x)=\frac{d}{dx}\int_{0}^{x}\left(\frac{1}{t^{2}}-2t+3\right)dt=\frac{1}{x^{2}}-2x+3$ .

Therefore,

$f'(6)=\frac{1}{6^{2}}-2(6)+3=\frac{1}{36}-12+3=\frac{1}{36}-9=\frac{1}{36}-\frac{324}{36}=-\frac{323}{36}$

Find the result:

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{e^x} t \ln t ; \mathrm{d}t$

$xe^{2x}$

$xe^{x+1}$

$xe^{x }+ e$

Explanation

Set . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = e^x$ , and by the chain rule,

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{e^x} t \ln t ; \mathrm{d}t$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{1}^{u} t \ln t ; \mathrm{d}t$

$= e^x \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{1}^{u} t \ln t ; \mathrm{d}t$

By the fundamental theorem of Calculus, the above can be rewritten as

$e^x \cdot u \ln u$

$= e^x \cdot e^x \ln e^x$

$= e^{2x} \cdot x = xe^{2x}$

Evaluate when $f(x)=\int_{0}^{x}(6t^{2}-3t+10)dt$ .

Explanation

Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_{0}^{x}(6t^{2}-3t+10)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(6t^{2}-3t+10)dt=6x^{2}-3x+10$ . Therefore, $f'(1)=6(1)^{2}-3(1)+10=6-3+10=3+10=13$ .

Fundamental Theorem of Calculus with Definite Integrals

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation