Integrals and the Fundamental Theorem: math test solutions, College – download the PDF

1 Antiderivatives / 3 pts

1. \(F(x)=x^5-x^3+\dfrac{x^2}{2}-2x+C\). (1.5 pt)
2. \(F(x)=2x^3+2x^2+C\); \(F(1)=4+C=3\) gives \(C=-1\), so \(F(x)=2x^3+2x^2-1\). (1.5 pt)

2 Riemann sums / 4 pts

\(\Delta x=0.5\).

1. Right endpoints 0.5, 1, 1.5, 2 give heights 1.25, 3, 5.25, 8: sum 17.5, so \(R_4=8.75\). (1.5 pt)
2. Left endpoints 0, 0.5, 1, 1.5 give heights 0, 1.25, 3, 5.25: sum 9.5, so \(L_4=4.75\). (1.5 pt)
3. \(\left[\dfrac{x^3}{3}+x^2\right]_0^2=\dfrac83+4=\dfrac{20}{3}\approx6.67\), and \(4.75<6.67<8.75\). (1 pt)

3 Properties / 3 pts

1. \(4+(-1)=3\). (1 pt)
2. \(-\int_3^7f=1\). (1 pt)
3. \(2(3)-5=1\). (1 pt)

4 Fundamental Theorem / 4 pts

1. \([2x^3-x^2]_0^2=(16-4)-0=12\). (1.5 pt)
2. \(F'(x)=\sqrt{x^2+9}\), so \(F'(4)=\sqrt{25}=5\). (1 pt)
3. Chain rule: \(G'(x)=\big((2x)^2+1\big)\cdot2=8x^2+2\), so \(G'(1)=10\). (1.5 pt)

5 Net change / 3 pts

1. \(\displaystyle\int_0^4(200+30t)\,dt=[200t+15t^2]_0^4=800+240=1{,}040\) gallons. (2 pts)
2. \(5{,}000-1{,}040=3{,}960\) gallons. (1 pt)

6 Substitution / 3 pts

1. Let \(u=x^2-4\), \(du=2x\,dx\): \(3\int u^3du=\dfrac{3u^4}{4}+C=\dfrac34(x^2-4)^4+C\). (1.5 pt)
2. Let \(u=x^3\), \(x^2dx=\tfrac13du\), limits 0 to 1: \(\tfrac13[e^u]_0^1=\dfrac{e-1}{3}\approx0.573\). (1.5 pt)