Calculus 2 Exam

184

310

remember width= b-a/n

Consider f(x) = x² + 1 on the domain [0, 4]. Find $L_f(4)$, the lower Riemann sum of f(x) on [0, 4] when the domain is partitioned into 4 equal intervals.

18

F(x) = sin (x) + 1

$F(x) = \int_{\pi}^{x} \sin(\theta)\,d\theta.$ Express the following as a function of x (without integral signs)

F(x) = -cos (x) -1

$G(x) = \int_{1}^{x} 3t^2\,dt.$ Express the following as a function of x (without integral signs)

x³-1

32

$$F(x) = \int_{1}^{x} e^{t^2+2}\,dt.$$

e^11

$$G(x) = \int_{5}^{x} \sqrt{t^3+1}\,dt.$$

3

remember when both limits have x, you use chain rule and. 2^17(4) -2^5

3x^7/3 / 7 + 3x^4/3 / 4 + C

find $\int \sqrt[3]{x}\,(x^2 - 2)\,dx$

$$\int \sqrt[3]{x}\,(x^2-2)\,dx = \frac{3}{10}x^{10/3} - \frac{3}{2}x^{4/3} + C$$

find $\int \sqrt{x}\,(x + 4)\,dx$

6

3

lim =0 ; Let epsilon be any number greater than zero. Choose delta to be equal to epsilon squared. This works because if 0< I x+3I < epsilon squared, then I the square root of x + 3 - 0I < epsilon. This works because for every epsilon greater than zero, there exists a delta greater than zero that proves the limit to be true.

Let x be any number greater than zero. Choose H to be 0.01/x. This works because we know H has to be greater than zero, and the quotient of two positive numbers is always positive.

y=0.1

1/10

epsilon, 1, 4, delta, 6, epsilon

square root of epsilon

a= 5, b=11

2 and 10

1y²sin(y)

3x² square root of (x^6) + 1

297

U= 1104 L=624