This problem applies the Fundamental Theorem of Calculus Part 1, which tells us that differentiating an integral with variable upper limit gives the integrand evaluated at that limit. For $H(x) = \int_{0}^{x}(\sin t + t^2),dt$, taking the derivative yields $H'(x) = \sin t + t^2$ evaluated at $t=x$. Therefore, $H'(x) = \sin x + x^2$, which is simply the integrand with $t$ replaced by $x$. Choice A ($\cos x + 2x$) might attract students who incorrectly differentiate each term of the integrand rather than evaluating it. To apply FTC Part 1, identify the variable upper limit and substitute it into the integrand—no additional differentiation needed.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function G(x) is defined as the integral from 4 to x of 1/(t ln t) dt, which accumulates the area under the curve of f(t) = 1/(t ln t) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative G'(x) equals the integrand evaluated at x, which is 1/(x ln x). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from 4 to x, but that actually computes G(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function g(x) is defined as the integral from 0 to x of 1/(1+cos t) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative g'(x) is simply the integrand evaluated at t = x, so g'(x) = 1/(1+cos x). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function f(x) is defined as the integral from 3 to x of (ln $t)^3$ dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative f'(x) is simply the integrand evaluated at t = x, so f'(x) = (ln $x)^3$. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function e(x) is defined as the integral from -2 to x of $t/sqrt(t^2$+5) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative e'(x) is simply the integrand evaluated at t = x, so e'(x) = $x/sqrt(x^2$+5). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function c(x) is defined as the integral from -4 to x of $1/(t-1)^4$ dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative c'(x) is simply the integrand evaluated at t = x, so c'(x) = $1/(x-1)^4$. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function Z(x) is defined as the integral from -1 to x of $(e^{2t}$ + $ln(1+t^2$)) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative Z'(x) is simply the integrand evaluated at t = x, so Z'(x) = $e^{2x}$ + $ln(1+x^2$). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function Y(x) is defined as the integral from 0 to x of (sin $t)/(t^2$+1) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative Y'(x) is simply the integrand evaluated at t = x, so Y'(x) = (sin $x)/(x^2$+1). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function V(x) is defined as the integral from 2 to x of ($\sqrt{t} + t^{1/3}$) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative V'(x) is simply the integrand evaluated at t = x, so V'(x) = $\sqrt{x} + x^{1/3}$. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function W(x) is defined as the integral from 1 to x of 1/(1+sqrt(t)) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative W'(x) is simply the integrand evaluated at t = x, so W'(x) = 1/(1+sqrt(x)). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.