The integral ∫−123π2xsin(πx)dx is equal to:
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Step-by-Step Solution
Step 1: Analyze the integrand and absolute value
The integrand is f(x)=π2xsin(πx). We need to evaluate the integral of its absolute value. The absolute value function changes the sign of negative values to positive. Therefore, we need to determine the intervals where f(x) is positive and where it is negative within the integration limits of x=−1 to x=23.
Step 2: Determine the sign of the integrand
We analyze the sign of xsin(πx) in different intervals. For x∈[−1,0], both x and sin(πx) are non-positive, so their product is non-negative. For x∈[0,1], both x and sin(πx) are non-negative, so their product is non-negative. For x∈[1,23], x is positive but sin(πx) is non-positive (since πx is in [π,23π]), so their product is non-positive. Thus, f(x)≥0 for x∈[−1,1] and f(x)≤0 for x∈[1,23].
Step 3: Split the integral
Based on the sign analysis, we split the integral into two parts. For x∈[−1,1], ∣f(x)∣=f(x), and for x∈[1,23], ∣f(x)∣=−f(x). We can factor out π2 from the integrals.
Step 4: Apply integration by parts
We use integration by parts for ∫xsin(πx)dx. Let u=x and dv=sin(πx)dx. Then du=dx and v=−π1cos(πx). Applying the formula ∫udv=uv−∫vdu, we get the indefinite integral.
Step 5: Evaluate the definite integrals
Now we substitute the limits of integration into the antiderivative. For the first integral from −1 to 1: (−π1cos(π)+π21sin(π))−(−π−1cos(−π)+π21sin(−π))=(π1+0)−(π1+0)=0. For the second integral from 1 to 23: (−π3/2cos(23π)+π21sin(23π))−(−π1cos(π)+π21sin(π))=(0−π21)−(π1+0)=−π21−π1. Combining these with the negative sign for the second integral and multiplying by π2, we get I=π2[0−(−π21−π1)]=π2[π21+π1]=1+π. Wait, let's recheck the first integral. [−πxcos(πx)+π21sin(πx)]−11=(−π1cos(π)+π21sin(π))−(−π−1cos(−π)+π21sin(−π))=(−π1(−1)+0)−(π1(−1)+0)=π1−(−π1)=π2. So the first part is π2(π2)=2π. The second part is −π2[(0−π21)−(π1+0)]=−π2[−π21−π1]=1+π. Adding them: 2π+1+π=3π+1. Let's re-evaluate the first integral carefully. I1=∫−11π2xsin(πx)dx. Since xsin(πx) is an even function ((−x)sin(−πx)=(−x)(−sin(πx))=xsin(πx)), I1=2π2∫01xsin(πx)dx. 2π2[−πxcos(πx)+π21sin(πx)]01=2π2[(−π1cos(π)+π21sin(π))−(0+0)]=2π2[−π1(−1)+0]=2π2[π1]=2π. This is correct. Now for the second integral: I2=∫123−π2xsin(πx)dx=−π2[−πxcos(πx)+π21sin(πx)]123. I2=−π2[(−π3/2cos(23π)+π21sin(23π))−(−π1cos(π)+π21sin(π))]. I2=−π2[(0+π21(−1))−(−π1(−1)+0)]. I2=−π2[−π21−π1]=1+π. Total integral I=I1+I2=2π+(1+π)=3π+1.