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Integrate the function $x^4+2yx^2-16x^2+12x-4yx+27-6y$ from $-3$ to $5$

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Solution

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Final Answer

$\frac{2624}{15}+\frac{64}{3}y$
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Step-by-step Solution

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Expand the integral $\int_{-3}^{5}\left(x^4+2yx^2-16x^2+12x-4yx+27-6y\right)dx$ into $7$ integrals using the sum rule for integrals, to then solve each integral separately

$\int_{-3}^{5} x^4dx+\int_{-3}^{5}2yx^2dx+\int_{-3}^{5}-16x^2dx+\int_{-3}^{5}12xdx+\int_{-3}^{5}-4yxdx+\int_{-3}^{5}27dx+\int_{-3}^{5}-6ydx$

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$\int_{-3}^{5} x^4dx+\int_{-3}^{5}2yx^2dx+\int_{-3}^{5}-16x^2dx+\int_{-3}^{5}12xdx+\int_{-3}^{5}-4yxdx+\int_{-3}^{5}27dx+\int_{-3}^{5}-6ydx$

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Learn how to solve differential calculus problems step by step online. Integrate the function x^4+2yx^2-16x^212x-4yx+27-6y from -3 to 5. Expand the integral \int_{-3}^{5}\left(x^4+2yx^2-16x^2+12x-4yx+27-6y\right)dx into 7 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int_{-3}^{5} x^4dx results in: \frac{3368}{5}. The integral \int_{-3}^{5}2yx^2dx results in: \frac{304}{3}y. The integral \int_{-3}^{5}-16x^2dx results in: -\frac{2432}{3}.

Final Answer

$\frac{2624}{15}+\frac{64}{3}y$

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Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of (x^4+2y)dx from -3 to 5 using partial fractionsSolve integral of (x^4+2y)dx from -3 to 5 using basic integralsSolve integral of (x^4+2y)dx from -3 to 5 using u-substitutionSolve integral of (x^4+2y)dx from -3 to 5 using integration by partsSolve integral of (x^4+2y)dx from -3 to 5 using trigonometric substitution

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(◻)
+
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◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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Main Topic: Differential Calculus

The derivative of a function of a real variable measures the sensitivity to change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). Derivatives are a fundamental tool of calculus.

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