Resolved: "Evaluate the following integrals. ∫−11w(w2+1)3dw"

Joan Thompson

Answered question

2021-12-10

Evaluate the following integrals.

\int_{- 1}^{1} w {(w^{2} + 1)}^{3} d w

sonorous9n

Beginner2021-12-11Added 34 answers

Step 1
To evaluate the integral:

\int_{- 1}^{1} w {(w^{2} + 1)}^{3} d w

Solution:
Property of definite integral states that:

\int_{- a}^{a} f (x) d x = 0

, if f(-x)=-f(x) (Odd function)

\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x

, if f(-x)=f(x) (Even function)
Here, we have

f (w) = w {(w^{2} + 1)}^{3}

f (- w) = - w {({(- w)}^{2} + 1)}^{3}

= - w {(w^{2} + 1)}^{3}

=-f(w)
Step 2
So, function is odd function.
Therefore, using property

\int_{- 1}^{1} w {(w^{2} + 1)}^{3} d w = 0

Hence, required answer is 0.

Pansdorfp6

Beginner2021-12-12Added 27 answers

\int_{- 1}^{1} w \cdot {(w^{2} + 1)}^{3} d w

\int w \cdot {(w^{2} + 1)}^{3} d w

\int \frac{p^{3}}{2} d p

\frac{1}{2} \cdot \int p^{3} d p

\frac{1}{2} \cdot \frac{p^{4}}{4}

\frac{1}{2} \cdot \frac{{(w^{2} + 1)}^{4}}{4}

Multiply

\frac{{(w^{2} + 1)}^{4}}{8}

Return the limits

\frac{{(w^{2} + 1)}^{4}}{8} ∣_{- 1}^{1}

\frac{{(1^{2} + 1)}^{4}}{8} - \frac{{({(- 1)}^{2} + 1)}^{4}}{8}

Calculate
Answer:
0

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