Question

The value of $\int_{-10}^{10}(0.0002x^{3}-0.3x+20)dx$ is equal to

• A. 423
• B. 400
• C. 378
• D. 410
• E. 390

Hint:

Calculus Tip: Always check for symmetry in the bounds (like $-a$ to $a$). If you spot it, immediately cross out any terms with an odd exponent ($x, x^3, \sin x$). You only have to integrate the even powers and constants!

Answer 1

The Correct Option is B

Concept:
For any odd function $f(x)$ where $f(-x) = -f(x)$, the definite integral over a symmetric interval $[-a, a]$ is always exactly zero: $\int_{-a}^a f(x)dx = 0$. This property allows us to instantly eliminate terms with odd powers of $x$.

Step 1: Split the integral into separate terms.
By the linearity of integrals, we can separate the components: $$I = \int_{-10}^{10} 0.0002x^{3} dx - \int_{-10}^{10} 0.3x dx + \int_{-10}^{10} 20 dx$$

Step 2: Identify and eliminate odd functions.
The functions $x^3$ and $x^1$ are odd functions. Therefore, their integrals over the symmetric interval $[-10, 10]$ evaluate to zero: $$\int_{-10}^{10} 0.0002x^{3} dx = 0$$ $$\int_{-10}^{10} 0.3x dx = 0$$

Step 3: Focus on the remaining even/constant term.
The only term that does not evaluate to zero is the constant: $$I = 0 - 0 + \int_{-10}^{10} 20 dx$$

Step 4: Integrate the constant.
The integral of a constant $k$ is $kx$: $$I = [20x]_{-10}^{10}$$

Step 5: Evaluate the definite integral.
Substitute the upper and lower limits: $$I = 20(10) - 20(-10)$$ $$I = 200 - (-200)$$ $$I = 400$$ Hence the correct answer is (B) 400.