Question

in exercises 15 - 24, evaluate the integral analytically by using the integral evaluation theorem (part 2 of the fundamental theorem 4). 15. $int_{-2}^{2}5dx$ 16. $int_{2}^{5}4xdx$ 17. $int_{0}^{pi/4}cos xdx$ 18. $int_{-1}^{1}(3x^{2}-4x + 7)dx$ 19. $int_{0}^{1}(8s^{3}-12s^{2}+5)ds$ 20. $int_{1}^{2}\frac{4}{x^{2}}dx$ 21. $int_{1}^{27}y^{-4/3}dy$ 22. $int_{1}^{4}\frac{dt}{tsqrt{t}}$ 23. $int_{0}^{pi/3}sec^{2}\theta d\theta$ 24. $int_{1}^{e}(1/x)dx$

Explanation:

Step1: Recall the power - rule for integration

The power - rule for integration is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), and $\int kdx=kx + C$ (where $k$ is a constant), $\int\cos xdx=\sin x + C$, $\int\sec^{2}x dx=\tan x + C$, $\int\frac{1}{x}dx=\ln|x|+C$.

Step2: Evaluate $\int_{-2}^{2}5dx$

Using the rule $\int kdx=kx + C$, we have $\int_{-2}^{2}5dx=5x\big|_{-2}^{2}$.
Substitute the upper and lower limits: $5\times2-5\times(-2)=10 + 10=20$.

Step3: Evaluate $\int_{2}^{5}4x dx$

Using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n+1}$, for $y = 4x$, $\int 4x dx=4\times\frac{x^{2}}{2}=2x^{2}+C$.
Then $\int_{2}^{5}4x dx=2x^{2}\big|_{2}^{5}=2\times5^{2}-2\times2^{2}=2\times(25 - 4)=2\times21 = 42$.

Step4: Evaluate $\int_{0}^{\frac{\pi}{4}}\cos xdx$

Using the rule $\int\cos xdx=\sin x + C$, we have $\int_{0}^{\frac{\pi}{4}}\cos xdx=\sin x\big|_{0}^{\frac{\pi}{4}}=\sin\frac{\pi}{4}-\sin0=\frac{\sqrt{2}}{2}-0=\frac{\sqrt{2}}{2}$.

Step5: Evaluate $\int_{-1}^{1}(3x^{2}-4x + 7)dx$

First, integrate term - by - term: $\int(3x^{2}-4x + 7)dx=3\times\frac{x^{3}}{3}-4\times\frac{x^{2}}{2}+7x=x^{3}-2x^{2}+7x + C$.
Then $\int_{-1}^{1}(3x^{2}-4x + 7)dx=(x^{3}-2x^{2}+7x)\big|_{-1}^{1}=(1^{3}-2\times1^{2}+7\times1)-((-1)^{3}-2\times(-1)^{2}+7\times(-1))=(1 - 2 + 7)-(-1-2 - 7)=6-(-10)=16$.

Step6: Evaluate $\int_{0}^{1}(8s^{3}-12s^{2}+5)ds$

Integrate term - by - term: $\int(8s^{3}-12s^{2}+5)ds=8\times\frac{s^{4}}{4}-12\times\frac{s^{3}}{3}+5s=2s^{4}-4s^{3}+5s + C$.
Then $\int_{0}^{1}(8s^{3}-12s^{2}+5)ds=(2s^{4}-4s^{3}+5s)\big|_{0}^{1}=2\times1^{4}-4\times1^{3}+5\times1-0=2 - 4 + 5=3$.

Step7: Evaluate $\int_{1}^{2}\frac{4}{x^{2}}dx$

Rewrite $\frac{4}{x^{2}}$ as $4x^{-2}$. Then $\int 4x^{-2}dx=4\times\frac{x^{-2 + 1}}{-2+1}=-4x^{-1}+C=-\frac{4}{x}+C$.
So $\int_{1}^{2}\frac{4}{x^{2}}dx=-\frac{4}{x}\big|_{1}^{2}=-\frac{4}{2}+\frac{4}{1}=-2 + 4=2$.

Step8: Evaluate $\int_{1}^{27}y^{-\frac{4}{3}}dy$

Using the power - rule $\int y^{n}dy=\frac{y^{n + 1}}{n+1}$, for $n=-\frac{4}{3}$, $\int y^{-\frac{4}{3}}dy=\frac{y^{-\frac{4}{3}+1}}{-\frac{4}{3}+1}=\frac{y^{-\frac{1}{3}}}{-\frac{1}{3}}=-3y^{-\frac{1}{3}}+C$.
Then $\int_{1}^{27}y^{-\frac{4}{3}}dy=-3y^{-\frac{1}{3}}\big|_{1}^{27}=-3\times27^{-\frac{1}{3}}+3\times1^{-\frac{1}{3}}=-3\times\frac{1}{3}+3=-1 + 3=2$.

Step9: Evaluate $\int_{1}^{4}\frac{dt}{t\sqrt{t}}$

Rewrite $\frac{1}{t\sqrt{t}}$ as $t^{-\frac{3}{2}}$. Then $\int t^{-\frac{3}{2}}dt=\frac{t^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}=\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}=-2t^{-\frac{1}{2}}+C$.
So $\int_{1}^{4}\frac{dt}{t\sqrt{t}}=-2t^{-\frac{1}{2}}\big|_{1}^{4}=-2\times4^{-\frac{1}{2}}+2\times1^{-\frac{1}{2}}=-2\times\frac{1}{2}+2=-1 + 2=1$.

Step10: Evaluate $\int_{0}^{\frac{\pi}{3}}\sec^{2}\theta d\theta$

Using the rule $\int\sec^{2}\theta d\theta=\tan\theta + C$, we have $\int_{0}^{\frac{\pi}{3}}\sec^{2}\theta d\theta=\tan\theta\big|_{0}^{\frac{\pi}{3}}=\tan\frac{\pi}{3}-\tan0=\sqrt{3}-0=\sqrt{3}$.

Step11: Evaluate $\int_{1}^{e}\frac{1}{x}dx$

Using the rule $\int\frac{1}{x}dx=\ln|x|+C$, we have $\int_{1}^{e}\frac{1}{x}dx=\ln x\big|_{1}^{e}=\ln e-\ln1=1-0=1$.

1. Answer: $20$
2. Answer: $42$
3. Answer: $\frac{\sqrt{2}}{2}$
4. Answer: $16$
5. Answer: $3$
6. Answer: $2$
7. Answer: $2$
8. Answer: $1$
9. Answer: $\sqrt{3}$
10. Answer: $1$

Answer:

Step1: Recall the power - rule for integration

The power - rule for integration is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), and $\int kdx=kx + C$ (where $k$ is a constant), $\int\cos xdx=\sin x + C$, $\int\sec^{2}x dx=\tan x + C$, $\int\frac{1}{x}dx=\ln|x|+C$.

Step2: Evaluate $\int_{-2}^{2}5dx$

Using the rule $\int kdx=kx + C$, we have $\int_{-2}^{2}5dx=5x\big|_{-2}^{2}$.
Substitute the upper and lower limits: $5\times2-5\times(-2)=10 + 10=20$.

Step3: Evaluate $\int_{2}^{5}4x dx$

Using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n+1}$, for $y = 4x$, $\int 4x dx=4\times\frac{x^{2}}{2}=2x^{2}+C$.
Then $\int_{2}^{5}4x dx=2x^{2}\big|_{2}^{5}=2\times5^{2}-2\times2^{2}=2\times(25 - 4)=2\times21 = 42$.

Step4: Evaluate $\int_{0}^{\frac{\pi}{4}}\cos xdx$

Using the rule $\int\cos xdx=\sin x + C$, we have $\int_{0}^{\frac{\pi}{4}}\cos xdx=\sin x\big|_{0}^{\frac{\pi}{4}}=\sin\frac{\pi}{4}-\sin0=\frac{\sqrt{2}}{2}-0=\frac{\sqrt{2}}{2}$.

Step5: Evaluate $\int_{-1}^{1}(3x^{2}-4x + 7)dx$

First, integrate term - by - term: $\int(3x^{2}-4x + 7)dx=3\times\frac{x^{3}}{3}-4\times\frac{x^{2}}{2}+7x=x^{3}-2x^{2}+7x + C$.
Then $\int_{-1}^{1}(3x^{2}-4x + 7)dx=(x^{3}-2x^{2}+7x)\big|_{-1}^{1}=(1^{3}-2\times1^{2}+7\times1)-((-1)^{3}-2\times(-1)^{2}+7\times(-1))=(1 - 2 + 7)-(-1-2 - 7)=6-(-10)=16$.

Step6: Evaluate $\int_{0}^{1}(8s^{3}-12s^{2}+5)ds$

Integrate term - by - term: $\int(8s^{3}-12s^{2}+5)ds=8\times\frac{s^{4}}{4}-12\times\frac{s^{3}}{3}+5s=2s^{4}-4s^{3}+5s + C$.
Then $\int_{0}^{1}(8s^{3}-12s^{2}+5)ds=(2s^{4}-4s^{3}+5s)\big|_{0}^{1}=2\times1^{4}-4\times1^{3}+5\times1-0=2 - 4 + 5=3$.

Step7: Evaluate $\int_{1}^{2}\frac{4}{x^{2}}dx$

Rewrite $\frac{4}{x^{2}}$ as $4x^{-2}$. Then $\int 4x^{-2}dx=4\times\frac{x^{-2 + 1}}{-2+1}=-4x^{-1}+C=-\frac{4}{x}+C$.
So $\int_{1}^{2}\frac{4}{x^{2}}dx=-\frac{4}{x}\big|_{1}^{2}=-\frac{4}{2}+\frac{4}{1}=-2 + 4=2$.

Step8: Evaluate $\int_{1}^{27}y^{-\frac{4}{3}}dy$

Using the power - rule $\int y^{n}dy=\frac{y^{n + 1}}{n+1}$, for $n=-\frac{4}{3}$, $\int y^{-\frac{4}{3}}dy=\frac{y^{-\frac{4}{3}+1}}{-\frac{4}{3}+1}=\frac{y^{-\frac{1}{3}}}{-\frac{1}{3}}=-3y^{-\frac{1}{3}}+C$.
Then $\int_{1}^{27}y^{-\frac{4}{3}}dy=-3y^{-\frac{1}{3}}\big|_{1}^{27}=-3\times27^{-\frac{1}{3}}+3\times1^{-\frac{1}{3}}=-3\times\frac{1}{3}+3=-1 + 3=2$.

Step9: Evaluate $\int_{1}^{4}\frac{dt}{t\sqrt{t}}$

Rewrite $\frac{1}{t\sqrt{t}}$ as $t^{-\frac{3}{2}}$. Then $\int t^{-\frac{3}{2}}dt=\frac{t^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}=\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}=-2t^{-\frac{1}{2}}+C$.
So $\int_{1}^{4}\frac{dt}{t\sqrt{t}}=-2t^{-\frac{1}{2}}\big|_{1}^{4}=-2\times4^{-\frac{1}{2}}+2\times1^{-\frac{1}{2}}=-2\times\frac{1}{2}+2=-1 + 2=1$.

Step10: Evaluate $\int_{0}^{\frac{\pi}{3}}\sec^{2}\theta d\theta$

Using the rule $\int\sec^{2}\theta d\theta=\tan\theta + C$, we have $\int_{0}^{\frac{\pi}{3}}\sec^{2}\theta d\theta=\tan\theta\big|_{0}^{\frac{\pi}{3}}=\tan\frac{\pi}{3}-\tan0=\sqrt{3}-0=\sqrt{3}$.

Step11: Evaluate $\int_{1}^{e}\frac{1}{x}dx$

Using the rule $\int\frac{1}{x}dx=\ln|x|+C$, we have $\int_{1}^{e}\frac{1}{x}dx=\ln x\big|_{1}^{e}=\ln e-\ln1=1-0=1$.

1. Answer: $20$
2. Answer: $42$
3. Answer: $\frac{\sqrt{2}}{2}$
4. Answer: $16$
5. Answer: $3$
6. Answer: $2$
7. Answer: $2$
8. Answer: $1$
9. Answer: $\sqrt{3}$
10. Answer: $1$