Approximate the area bounded by $$ y=\sin (2 x) $$ (2) and $$ y=0.2 x+0.3 \quad 0 \leq x \leq \pi $$ [Answer in 2 decimal places]

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To approximate the area bounded by the curves $$ y=\sin (2x) $$ and $$ y=0.2x+0.3 $$ from $$ x=0 $$ to $$ x=\pi $$, we can use numerical integration.

The area bounded by the two curves can be calculated using the definite integral:

$$ \text{Area} = \int_{0}^{\pi} \left( \sin(2x) - (0.2x + 0.3) \right) dx $$

Let's calculate this integral numerically.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{\pi } \left(\sin\left(2x\right)-\left(0.2x+0.3\right)\right) dx$$
- step1: Remove the parentheses:
  $$\int_{0}^{\pi } \left(\sin\left(2x\right)-0.2x-0.3\right) dx$$
- step2: Evaluate the integral:
  $$\int \left(\sin\left(2x\right)-0.2x-0.3\right) dx$$
- step3: Use properties of integrals:
  $$\int \sin\left(2x\right) dx+\int -0.2x dx+\int -0.3 dx$$
- step4: Evaluate the integral:
  $$-\frac{1}{2}\cos\left(2x\right)+\int -0.2x dx+\int -0.3 dx$$
- step5: Evaluate the integral:
  $$-\frac{1}{2}\cos\left(2x\right)-\frac{x^{2}}{10}+\int -0.3 dx$$
- step6: Evaluate the integral:
  $$-\frac{1}{2}\cos\left(2x\right)-\frac{x^{2}}{10}-0.3x$$
- step7: Return the limits:
  $$\left(-\frac{1}{2}\cos\left(2x\right)-\frac{x^{2}}{10}-0.3x\right)\bigg |_{0}^{\pi }$$
- step8: Calculate the value:
  $$-\frac{\pi ^{2}}{10}-0.3\pi $$
- step9: Calculate:
  $$-\frac{\pi ^{2}+3\pi }{10}$$
The approximate area bounded by the curves $$ y=\sin (2x) $$ and $$ y=0.2x+0.3 $$ from $$ x=0 $$ to $$ x=\pi $$ is approximately -1.929438.

Since the area is negative, it means that the curve $$ y=\sin (2x) $$ is below the curve $$ y=0.2x+0.3 $$ in the given interval.

Therefore, the area bounded by the two curves is approximately 1.929438 square units.
The area bounded by $$ y=\sin (2x) $$ and $$ y=0.2x+0.3 $$ from $$ x=0 $$ to $$ x=\pi $$ is approximately 1.93 square units.

Approximate the area bounded by
(2) and

[Answer in 2 decimal places]

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Approximate the area bounded by (2) and [Answer in 2 decimal places]