# How to evaluate $\int_{-10}^{10}(8x^9+5x^5)\,dx$ using the fundamental theorem of calculus?

Can you somebody help me to solve this question?

Evaluate the intergral using the fundamental theorem of calculus.
$$\int_{-10}^{10}(8x^9+5x^5)\,dx$$

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First step: find an antiderivative of $8x^9+5x^5$. –  David Mitra Feb 10 '13 at 19:08
Note - you are integrating an odd function between limits which are symmetric about zero - if the function is integrable there is only ever going to be one answer. This is a comment rather than an answer to the question as it does not use FTC. However the observation is worth noting as it is occasionally useful. –  Mark Bennet Feb 10 '13 at 19:23

$\int_{-10}^{10}(8x^9+5x^5)dx=\int_{-10}^{10}8x^9dx+\int_{-10}^{10}5x^5dx=8\int_{-10}^{10}x^9dx+5\int_{-10}^{10}x^5)dx=(8\cdot\frac{x^{10}}{10}+5\cdot\frac{x^{6}}{6})|_{-10}^{10}=\frac{8}{10}(10^{10}-(-10)^{10})+\frac{5}{6}(10^6-(-10)^6)=\frac{8}{10}(10^{10}-10^{10})+\frac{5}{6}(10^6-10^6)=0$

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Maybe words and/or multiple lines would help –  andybenji Feb 10 '13 at 19:14
Organization is a good thing, I agree. Its ugly and difficult to follow. But I trust that is some respondents' intentions. –  CogitoErgoCogitoSum Feb 10 '13 at 19:54

(I will take this rather deliberately, you can certainly do it much faster.)

By the fundamental theorem of calculus, we have:

$$\int_a^b f(x) \mathrm dx = F(b) - F(a)$$

Where: $$F(x) = \int f(x) \mathrm dx$$

So:

$$\int 8x^9 + 5x^5 \mathrm dx$$

Break it into two parts:

$$\int 8x^9 \mathrm dx + \int 5x^5 \mathrm dx$$

Move the constant outside of the integral:

$$8\int x^9 \mathrm dx + 5\int x^5 \mathrm dx$$

The power rule for integration gives: $$8 \cdot \frac{x^{10}}{10} + C_1 + 5 \cdot \frac{x^6}{6} + C_2$$ $$\frac{8x^{10}}{10} + \frac{5x^6}{6} + C_1 + C_2$$

Merge the constants of integration: $$\frac{8x^{10}}{10} + \frac{5x^6}{6} + C$$

So:

$$\int_{-10}^{10} f(x) \mathrm dx = \frac{8 \cdot 10^{10}}{10} + \frac{5 \cdot 10^6}{6} + C - \left(\frac{8 \cdot (-10)^{10}}{10} + \frac{5 \cdot (-10)^6}{6} + C\right)$$

Note that the substitution $q = \frac{8 \cdot 10^{10}}{10} + \frac{5 \cdot 10^6}{6} + C$ makes it:

$$\int_{-10}^{10} f(x) \mathrm dx = q - q = 0$$

This holds for any odd function $g(x)$. Given:

$$g(-x) = -g(x)$$

We can show that if $G(x) = \int g(x) \mathrm dx$ (and $G(x)$ exists), then it is an even function. So:

$$\int_{-a}^a g(x) \mathrm dx = G(a) - G(-a)$$ Since $G(x)$ is odd, $G(a) = G(-a)$, so the integral is equal to $0$.

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There is no reason to go through the process of integration. You can easily see that the expression $y=8x^9 +5x^5$ is a polynomial with only odd-degree terms, and there is no constant term. This polynomial is symmetric about the origin. The integral is taken from -10 to +10, equidistant on either side of the origin. There will be equal areas above as below the x-axis and the total sum of area will be zero.

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