This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to 4 of √x dx equals F(4) - F(0), where F is $(2/3)x^{3/2}$. F(4)=(2/3)(8)=16/3, F(0)=0, so 16/3 - 0=16/3. This gives the area. A tempting distractor is choice A, F(0) - F(4)=0 - 16/3=-16/3, reversing the sign. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from -2 to 1 of $3x^2$ dx equals W(1) - W(-2), where W is $x^3$. W(1)=1, W(-2)=-8, so 1 - (-8)=9. Note the limits: upper 1, lower -2. A tempting distractor is choice B, W(-2) - W(1)=-8-1=-9, reversing the order. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 1 to 4 of v(t) dt equals V(4) - V(1), where V is the given antiderivative. To compute this, evaluate V at the upper limit 4 and subtract V at the lower limit 1. This direct substitution yields the net change in the antiderivative over the interval. A tempting distractor is choice B, V(1) - V(4), which reverses the order and results in the negative of the correct integral. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 1 to 9 of $4\sqrt{x} , dx$ equals $A(9) - A(1)$, where $A(x) = \dfrac{8}{3} x^{3/2}$. Evaluate $A(9) = \dfrac{8}{3} \times 27 = 72$, $A(1) = \dfrac{8}{3} \times 1 = \dfrac{8}{3}$, so $72 - \dfrac{8}{3} = \dfrac{216 - 8}{3} = \dfrac{208}{3}$. This gives the work done by the force. A tempting distractor is choice A, $A(1) - A(9)$, which negates the correct value. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to π/3 of cos x dx equals F(π/3) - F(0), where F is sin x. Compute sin(π/3) which is √3/2, and subtract sin(0) which is 0. This results in √3/2 - 0 = √3/2, the area under the curve. A tempting distractor is choice A, sin 0 - sin(π/3), which reverses subtraction and gives a negative result. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from -1 to 3 of (6x-5) dx equals G(3) - G(-1), where G is $3x^2$ - 5x. Compute G(3) = 3(9) - 5(3) = 12, and G(-1) = 3(1) - 5(-1) = 8, so 12 - 8 = 4. This calculation gives the correct integral value. A tempting distractor is choice A, G(-1) - G(3), which reverses the order and yields -4. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to π/6 of $sec^2$ x dx equals H(π/6) - H(0), where H is tan x. Evaluate tan(π/6) = 1/√3, and tan(0) = 0, so 1/√3 - 0. This provides the exact value of the integral. A tempting distractor is choice B, tan 0 - tan(π/6), which switches subtraction and gives a negative. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

This problem requires applying the Fundamental Theorem of Calculus Part 2, which states that if F'(x) = f(x), then ∫[a to b] f(x)dx = F(b) - F(a). Since we're given that F'(x) = f(x) and F(x) = x³ - 4x, we can directly apply the theorem. To evaluate ∫[-1 to 2] f(x)dx, we compute F(2) - F(-1). We find F(2) = 2³ - 4(2) = 8 - 8 = 0 and F(-1) = (-1)³ - 4(-1) = -1 + 4 = 3, giving us F(2) - F(-1) = 0 - 3 = -3. A common error would be reversing the order to get F(-1) - F(2) = 3, which incorrectly switches the bounds. Remember: when using FTC Part 2, always evaluate the antiderivative at the upper bound minus the antiderivative at the lower bound.

This problem requires applying the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral using an antiderivative. According to FTC Part 2, the integral from 0 to 2 of $2e^t$ dt equals R(2) - R(0), where R is $2e^t$. Evaluate R at 2 to get $2e^2$, then subtract R at 0 which is $2e^0$ = 2. This yields $2e^2$ - 2, the net accumulation. A tempting distractor is choice A, $2e^0$ - $2e^2$, which inverts the limits and negates the integral. Always remember to evaluate the antiderivative at the upper limit minus the value at the lower limit to find the definite integral.

This question applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using a known antiderivative. Given that H'(x) = h(x) and H(x) = sin x + 2x, we can find ∫[0 to π] h(x)dx = H(π) - H(0). Evaluating at the bounds: H(π) = sin(π) + 2π = 0 + 2π = 2π and H(0) = sin(0) + 2(0) = 0 + 0 = 0. Thus, ∫[0 to π] h(x)dx = H(π) - H(0) = 2π - 0 = 2π. The distractor H(0) - H(π) = -2π incorrectly reverses the evaluation order, yielding a negative result when the actual integral is positive. To correctly apply FTC Part 2, always evaluate F(upper bound) - F(lower bound), maintaining the proper order to get the correct sign.