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PROBLEM

4. 011xdx=\int_{0}^{11-x} d x=
(a) 12ln2\frac{1}{2} \ln 2
(b) 2ln2+12 \ln 2+1
(c) 2ln212 \ln 2-1
(d) 12ln21\frac{1}{2} \ln 2-1

STEP 1

1. The integral to solve is 0111+xdx\int_{0}^{1} \frac{1}{1+x} \, dx.
2. This is a definite integral involving a natural logarithm function.

STEP 2

1. Identify the integral form.
2. Evaluate the integral.
3. Apply the limits of integration.
4. Simplify the result.

STEP 3

Recognize that the integral 11+xdx\int \frac{1}{1+x} \, dx is the natural logarithm function:
11+xdx=ln1+x+C \int \frac{1}{1+x} \, dx = \ln|1+x| + C

STEP 4

Evaluate the integral from 0 to 1:
0111+xdx=[ln1+x]01 \int_{0}^{1} \frac{1}{1+x} \, dx = \left[ \ln|1+x| \right]_{0}^{1}

STEP 5

Apply the limits of integration:
[ln1+x]01=ln1+1ln1+0 \left[ \ln|1+x| \right]_{0}^{1} = \ln|1+1| - \ln|1+0| =ln2ln1 = \ln 2 - \ln 1

SOLUTION

Simplify the result:
Since ln1=0\ln 1 = 0, we have:
ln20=ln2 \ln 2 - 0 = \ln 2 The value of the integral is:
ln2 \boxed{\ln 2} However, none of the provided options match ln2\ln 2 directly. Please verify the problem statement or options.

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