solve this---**Question Stem:** Find the area of the shaded region. **Chart Description:** * **Type:** A graph plotted on a 2D Cartesian coordinate system. * **Axes:** * A horizontal X-axis with an arrow pointing to the right. * A vertical Y-axis with an arrow pointing upwards. * The intersection of the axes is the origin (implied). * Tick marks are present on both axes, but no scale values are explicitly labeled other than the points. * **Curves:** * A curve labeled y(x) = 1 - x^2. This is a downward-opening parabola with its vertex at (0, 1). It intersects the x-axis at x = 1 and x = -1 (the point (1,0) is labeled). * A curve labeled y(x) = 2^x. This is an exponential growth curve passing through the point (0, 1). * **Points:** * A point labeled (0,1). This point is the intersection of both curves and also the vertex of the parabola and the y-intercept of the exponential function. * A point labeled (1,0). This point is an x-intercept of the parabola. * **Shaded Region:** * The shaded region is bounded above by the curve y(x) = 2^x and below by the curve y(x) = 1 - x^2. * The shaded region is also bounded on the left by the Y-axis (where x = 0) and on the right by a vertical dashed line extending from the point (1,0) upwards, which is the line x = 1. * **Other Labels:** The equations of the curves are explicitly labeled: y(x) = 2^x and y(x) = 1 - x^2. The coordinate axes are labeled 'x' and 'y'. **Mathematical Formulas/Equations:** y(x) = 2^x y(x) = 1 - x^2 **Options:** (No options are provided in the image). **Other Relevant Text:** (None other than the question stem and labels on the graph).

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We need to find the area of the shaded region between two curves: y equals 2 to the x, and y equals 1 minus x squared, from x equals 0 to x equals 1. The curves intersect at point (0,1), and our region is bounded by the vertical line x equals 1. To find the area between two curves, we use the formula: integral from a to b of the absolute value of f of x minus g of x. Since 2 to the x is greater than or equal to 1 minus x squared in the interval from 0 to 1, we can write the area as the integral from 0 to 1 of 2 to the x minus the quantity 1 minus x squared. This simplifies to the integral of 2 to the x minus 1 plus x squared. We need to find the area of the shaded region bounded by two curves: y equals 2 to the x, and y equals 1 minus x squared. The region is bounded from x equals 0 to x equals 1, where both curves intersect at the point (0,1). The exponential curve is above the parabola in this interval. To find the area between two curves, we use the integral formula: A equals the integral from a to b of f of x minus g of x dx, where f of x is the upper curve and g of x is the lower curve. In our case, 2 to the x is above 1 minus x squared in the interval from 0 to 1, so our integral becomes the integral from 0 to 1 of 2 to the x minus 1 minus x squared dx, which simplifies to 2 to the x minus 1 plus x squared. Now we need to find the antiderivatives of each term. The antiderivative of 2 to the x is 2 to the x divided by the natural logarithm of 2. The antiderivative of negative 1 is negative x. The antiderivative of x squared is x cubed divided by 3. Combining these, our integral becomes 2 to the x over ln 2 minus x plus x cubed over 3, evaluated from 0 to 1. Now we evaluate the definite integral using the fundamental theorem of calculus. At x equals 1, we get 2 over ln 2 minus 1 plus 1 over 3. At x equals 0, we get 1 over ln 2 minus 0 plus 0, which is just 1 over ln 2. Subtracting the second from the first, we get 2 over ln 2 minus 1 over ln 2 minus 1 plus 1 over 3, which simplifies to 1 over ln 2 minus 2 over 3. The final answer is the area equals 1 over ln 2 minus 2 over 3. Converting to decimal form, this is approximately 1.4427 minus 0.6667, which equals approximately 0.776 square units. This represents the exact area of the shaded region bounded by the exponential curve y equals 2 to the x and the parabola y equals 1 minus x squared from x equals 0 to x equals 1. Now we evaluate each definite integral step by step using the fundamental theorem of calculus. First, for the exponential term: 2 to the x over ln 2 evaluated from 0 to 1 equals 2 over ln 2 minus 1 over ln 2, which equals 1 over ln 2. Second, for the constant term: x evaluated from 0 to 1 equals 1 minus 0, which is 1. Third, for the polynomial term: x cubed over 3 evaluated from 0 to 1 equals 1 over 3 minus 0, which is 1 over 3. Combining all terms, we get A equals 1 over ln 2 minus 1 plus 1 over 3. Now we combine all our results to get the final answer. We have A equals 1 over ln 2 minus 1 plus 1 over 3. Converting to a common denominator, this becomes 1 over ln 2 minus 2 over 3. For the numerical calculation, ln 2 is approximately 0.693, so 1 over ln 2 is approximately 1.443. Therefore, A is approximately 1.443 minus 0.667, which equals approximately 0.776. The exact answer is 1 over ln 2 minus 2 over 3, and the approximate decimal value is 0.776 square units. This represents the area of the shaded region between the two curves.