Symmetry in integrals Use symmetry to evaluate the following i...

Question

Symmetry in integrals Use symmetry to evaluate the following integrals.

∫²₋₂ (x² + x³) dx

Accepted Answer

Welcome back, everyone. Evaluate the integral from -3 to 3 of X to the power of 4 plus x to the power of 5 DX using symmetry. A 0 B 243 divided by 5, C 486 divided by 5. D 1701 divided by 10. So for this integral because we have a sum, we can split it into two integrals. The first one is going to be in the integral from -3 to 3 of X to the power of 4DX. And the second one is going to be the integral from -3 to 3 of X to the power of 5DX. Now, let's evaluate the symmetry of each function, starting with X the power of 4. We can label this as our function F of X, and then X the power of 5. Well, let's suppose that it is G of X. First of all, let's evaluate F of negative X. This is going to be equal to negative X to the power of 4, which is X to the power of 4 because we're raising a negative number to the power of 4 and 4 is an even number, right? So we can essentially erase -1 to the power of 4 to get rid of the negative sign, and essentially this means that X to the power of 4 is an even function, right? It has an even exponent. Now I'm sorry, of negative X is going to be equal to negative X or raise to the power of 5, and this can be written as negative X or raise the power of 5, which means that the function is odd. We have an odd exponent, and whenever we raise a negative number to an odd exponent, we get a negative number. Now, whenever we integrate an odd function from negative A to A, where A is a number, we can state that the integral is equal to 0. This is how we use symmetry. Now for an even function, we can rewrite it as the integral from 0 up to the upper bound of 3 of F of XDX in this case the integral of X the power of 4DX. And because we are replacing our lower bound. With 0 instead of the original negative 3, we have to multiply that integral by 2. So we're going to use this property. This is going to be equal to 2. Now the integral of X to the power of 4 is X to the power of 5 divided by 5 using the power rule, and we are evaluating the result from 0 to 3. So we're using this property for the integrals of even functions. Instead of the original bounds, we are replacing the lower bound with 0 and multiplying the resultant integral by 2. So we get 2/5. A to the power of 5 from 0 to 3. Now applying the bounds, we get 2 5s multiplied by. 3 raise the power of 5 minus 0 raises the power of 5, which is 2 multiplied by 3 to the power of 5 divided by 5. And this is equal to 486 divided by 5, which corresponds to the answer choice C. Thank you for watching.

Symmetry in integrals Use symmetry to evaluate the following integrals.
∫²₋₂ (x² + x³) dx

Key Concepts

Symmetry in Functions

Definite Integrals

Properties of Integrals

Watch next