Find the interval on which the curve of y = ∫[0,x] 1/((1+t+t^2) dt is concave up.

Answer:     the interval on which the curve is concave up is  (-∞, -1/2)======    we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x    let the function f denote be the anti-derivative of the inside expression, i.e. a function such that .then use the fundamental theorem of calculus:     differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function    find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x    where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation    1 + x + x² = 0results in a discriminant of      meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get      so the since the 2nd derivative of the curve is positive on the interval  (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get      so the since the 2nd derivative of the curve is positive on the interval  (-∞, -1/2), the  curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is  (-∞, -1/2)

Pressure and heat that's what i think( don't report if wrong)

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