Compute integral: $\int_0^4 80\,dx - 0.5 \int_0^4 x^2\,dx = [80x]_0^4 - 0.5 \left[rac{x^3}{3}

["Compute the Integral Step-by-Step: A Guide to Evaluating $\int_0^4 (80 - 0.5x^2),dx$", "Solving definite integrals efficiently is essential in calculus and applied mathematics. One classic expression often encountered is:", "$$\n\int_0^4 \left(80 - 0.5x^2\right) dx\n$$", "Understanding how to compute this integral step-by-step helps build foundational skills for engineering, physics, and economics applications. In this article, we’ll evaluate this integral using fundamental integration techniques and verify the result with standard antiderivative notation, showing how to compute definite integrals from first principles.", "---", "### Understanding the Integral", "We begin with:", "$$\n\int_0^4 \left(80 - 0.5x^2\right) dx\n$$", "This represents the net area under the curve $ y = 80 - 0.5x^2 $ from $ x = 0 $ to $ x = 4 $. To evaluate it, we break the integral into two parts:", "$$\n\int_0^4 80, dx - \int_0^4 0.5x^2, dx\n$$", "---", "### Step 1: Compute $\int_0^4 80,dx$", "The integral of a constant $ c = 80 $ over $[0, 4]$ is:", "$$\n\int_0^4 80,dx = 80 \cdot (4 - 0) = 320\n$$", "---", "### Step 2: Compute $\int_0^4 0.5x^2,dx$", "Factor out the constant $ 0.5 $:", "$$\n\int_0^4 0.5x^2,dx = 0.5 \int_0^4 x^2,dx\n$$", "Now recall the standard antiderivative:", "$$\n\int x^2,dx = \frac{x^3}{3} + C\n$$", "Apply this:", "$$\n0.5 \int_0^4 x^2,dx = 0.5 \left[ \frac{x^3}{3} \right]_0^4 = 0.5 \left( \frac{4^3}{3} - 0 \right) = 0.5 \cdot \frac{64}{3} = \frac{32}{3}\n$$", "---", "### Step 3: Combine Results", "Now substitute both results back:", "$$\n\int_0^4 \left(80 - 0.5x^2\right) dx = 320 - \frac{32}{3}\n$$", "To write as a single expression (matching standard antiderivative notation):", "$$\n\boxed{ \int_0^4 \left(80 - 0.5x^2\right) dx = \left[ 80x \right]_0^4 - 0.5 \left[ \frac{x^3}{3} \right]_0^4 }\n$$", "This confirms our computation.", "---", "### Final Calculation (Optional Numerical Value)", "For practical purposes, compute:", "$$\n320 - \frac{32}{3} = \frac{960 - 32}{3} = \frac{928}{3} \approx 309.33\n$$", "---", "### Why This Integration Shell Matters", "This expression exemplifies:", "- Decomposing complex expressions into simpler integrals\n- Using basic antiderivatives\n- Evaluating definite integrals over a finite interval\n- Applying linearity of integration: $\int (a + b,f(x)),dx = a\int,dx + b\int,f(x),dx$", "Mastering such indefinite integrals prepares students and professionals alike for evaluating physical quantities, computing energy, work, or center of mass in applied disciplines.", "---", "Summary:\n$$\n\int_0^4 \left(80 - 0.5x^2\right) dx = \boxed{ \left[ 80x \right]_0^4 - 0.5 \left[ \frac{x^3}{3} \right]_0^4 } = 320 - \frac{32}{3} = \frac{928}{3}\n$$", "Use this framework whenever computing integrals—break, integrate, evaluate at bounds—ensuring accuracy in your mathematical solutions.", "---", "Keywords: Compute integral, definite integral, integration steps, calculator integral $ \int_0^4 (80 - 0.5x^2),dx $, antiderivative evaluation, applies to physics, engineering math."]