Question Evaluate the integral 2n22+C,C∈R Evaluate ∫4n2×1dn2Multiply the terms ∫4n2dn2Use the property of integral ∫kf(x)dx=k∫f(x)dx 4×∫n2dn2Use the property of integral ∫xndx=n+1xn+1 4×1+1n21+1Simplify More Steps Evaluate 1+1n21+1Add the numbers 1+1n22Add the numbers 2n22 4×2n22Cancel out the common factor 2 2n22Solution 2n22+C,C∈R Show Solution