Question
Evaluate the integral
34x2−4x+C,C∈R
Evaluate
∫2(2x−3)×32dx
Multiply the terms
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Multiply the terms
2(2x−3)×32
Multiply the terms
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Evaluate
2×32
Multiply the numbers
32×2
Multiply the numbers
34
34(2x−3)
Apply the distributive property
34×2x−34×3
Multiply the numbers
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Evaluate
34×2
Multiply the numbers
34×2
Multiply the numbers
38
38x−34×3
Multiply the numbers
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Evaluate
34×3
Reduce the numbers
4×1
Simplify
4
38x−4
∫(38x−4)dx
Rewrite the expression
∫34(2x−3)dx
Use the property of integral ∫kf(x)dx=k∫f(x)dx
34×∫(2x−3)dx
Use the property of integral ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx
34(∫2xdx+∫−3dx)
Calculate
34×∫2xdx+34×∫−3dx
Evaluate the integral
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Evaluate
34×∫2xdx
Use the property of integral ∫kf(x)dx=k∫f(x)dx
34×2×∫xdx
Multiply the numbers
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Evaluate
34×2
Multiply the numbers
34×2
Multiply the numbers
38
38×∫xdx
Use the property of integral ∫xndx=n+1xn+1
38×1+1x1+1
Simplify
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Evaluate
1+1x1+1
Add the numbers
1+1x2
Add the numbers
2x2
38×2x2
Cancel out the common factor 2
34x2
Multiply the terms
34x2
34x2+34×∫−3dx
Evaluate the integral
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Evaluate
34×∫−3dx
Use the property of integral ∫kdx=kx
34(−3x)
Multiply the numbers
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Evaluate
34(−3)
Multiplying or dividing an odd number of negative terms equals a negative
−34×3
Reduce the numbers
−4×1
Simplify
−4
−4x
34x2−4x
Solution
34x2−4x+C,C∈R
Show Solution
