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2 years ago

5

Could you help me solve?

1 answer:

Ilya [14]2 years ago

8 0

Answer:

1.14200738982×10^26

Step-by-step explanation:

Substitution can make this integral much easier to evaluate.

<h3>Substitution</h3>

Let u = 7x² -x. Then du = (14x -1)dx. The limits on x become different limits for u:

for x = 1: u = 7(1²) -1 = 6

for x = 3: u = 7(3²) -3 = 60

<h3>Integral</h3>

$\displaystyle\int_1^3{(14x-1)e^{(7x^2-x)}}\,dx=\int_6^{60}{e^u}\,du=\left.e^u\right|^{60}_6\\\\=e^{60}-e^6\approx e^{60}\approx\boxed{1.14200738982\times10^{26}}$

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3 0

3 years ago

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Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)

zloy xaker [14]

Answer:

A)

$k=0$

B)

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C)

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Step-by-step explanation:

We are given the function:

$\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}$

A)

Given that h(1) = 20, we want to find <em>k</em>.

h(1) = 20 means that <em>h</em>(x) = 20 when <em>x</em> = 1. Substitute:

$\displaystyle (20) = 20e^{k(1)}$

Simplify:

$1= e^k$

Anything raised to zero (except for zero) is one. Therefore:

$k=0$

B)

Given that h(1) = 40, we want to find 2<em>k</em> + 1.

Likewise, this means that <em>h</em>(x) = 40 when <em>x</em> = 1. Substitute:

$\displaystyle (40) = 20e^{k(1)}$

Simplify:

$\displaystyle 2 = e^{k}$

We can take the natural log of both sides:

$\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}$

By definition, ln(e) = 1. Hence:

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$2k+1 = 2\ln 2+ 1 \approx 2.3863$

C)

Given that h(1) = 10, we want to find <em>k</em> - 3.

Again, this meas that <em>h</em>(x) = 10 when <em>x</em> = 1. Substitute:

$\displaystyle (10) = 20e^{k(1)}$

Simplfy:

$\displaystyle \frac{1}{2} = e^k$

Take the natural log of both sides:

$\displaystyle \ln \frac{1}{2} = k\ln e$

Therefore:

$\displaystyle k = \ln \frac{1}{2}$

Therefore:

$\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931$

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3 years ago

What's the answer to 4n - 9 = -9

aksik [14]

Hello!

You solve this algebraically

4n - 9 = -9

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3 years ago

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Firdavs [7]

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3 years ago

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