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2 years ago

If the triangle ABC=JKL and ACB=12,KL=6cm,and AC=8.5cm what is BC

Mathematics

1 answer:

Illusion [34]2 years ago

5 0

BC is equal to KL so there for BC is 6 cm

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Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral $\int _0^3\:3-\frac{x}{3e^x}dx$ . We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx$

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Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{d}{dx}(x)$

$\frac{du}{dx}=1$

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$du=dx$

$v'=\frac{1}{e^x}$

$v=-\frac{1}{e^x}$

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

Therefore, the value of the given integral would be 8.733046.

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