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Question

If fx=5x8+7x6x2+1+2x72dx,x0,f0=0 and f1=1k, then the value of k is


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Solution

Step 1. Simplify the given function.

In the integral of the function fx=5x8+7x6x2+1+2x72dx divide the numerator and denominator by x14.

fx=5x8+7x6x14x2+1+2x7x72dx=5x-6+7x-8x-5+x-1+22dx

Now, let 1x-5+x-7+2=t, then 5x-6+7x8x-5+x-1+22dx=dt. So the integral becomes

5x-6+7x-8x-5+x-1+22dx=dt=t+C=1x-5+x-7+2+Ct=1x-5+x-7+2

So the function is fx=1x-5+x-7+2+C.

Step 2. Find the value of C.

The function fx=1x-5+x-7+2+C can be simplified as:

fx=x7x2+1+2x7+C

It is given that f0=0, so for x=0, the value of can be found as:

f0=0702+1+2×07+C0=0+C0=C

So, the function is fx=x7x2+1+2x7.

Step 3. Find the value of k.

For the function fx=x7x2+1+2x7, the value of f1 is given as:

f1=1712+1+2×17=11+1+2=14

But it is given that f1=1k, so the value of k is 4.

Hence, the value of k is 4.


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