Let ∫x3sinxdx=g(x)+C, where C is the constant of integration. If 8(g(2π)+g′(2π))=απ3+βπ2+γ, α,β,γ∈Z, then α+β−γ equals:
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Step-by-Step Solution
Step 1: Identify g(x) and g'(x)
The problem states that g(x)+C=∫x3sinxdx. By the Fundamental Theorem of Calculus, the derivative of an integral of a function is the function itself. Therefore, g'(x) is simply the integrand x3sinx.
Step 2: Evaluate g'(π/2)
Now we substitute x=2π into the expression for g'(x). We know that sin(2π)=1.
Step 3: Integrate g(x) using Integration by Parts
To find g(x), we need to evaluate the integral ∫x3sinxdx. We use integration by parts repeatedly, applying the formula ∫udv=uv−∫vdu. We choose u as the polynomial term and dv as the trigonometric term.
Step 4: Evaluate g(π/2)
Now we substitute x=2π into the expression for g(x). We use the values cos(2π)=0 and sin(2π)=1.
Step 5: Calculate the given expression
Substitute the calculated values of g(2π) and g′(2π) into the given expression 8(g(2π)+g′(2π)) and simplify.
Step 6: Determine α, β, γ and calculate α + β - γ
By comparing the simplified expression with απ3+βπ2+γ, we can identify the integer values of α, β, and γ. Then, we calculate α+β−γ.