Calculus/Fundamental Theorem of Calculus/Solutions

5. Evaluate $$\int\limits_1^4 2x^3+\frac{1}{3x^2}-4\sqrt x+5dx$$.$$\begin{aligned} \int\limits_1^4 2x^3+\frac{1}{3x^2}-4\sqrt x+5dx &=\int\limits_1^4 2x^3dx+\int\limits_1^4\frac1{3x^2}dx-\int\limits_1^44\sqrt xdx+\int\limits_1^45dx \\ &= 2\int\limits_1^4 x^3dx+\frac13\int\limits_1^4\frac1{x^2}dx-4\int\limits_1^4\sqrt xdx+5\int\limits_1^4dx \\ &=\left. \frac{x^4}{2}-\frac{1}{3x}-\frac{8x^{\frac32}}{3}+5x \right|^4_1 \\ &=\left[ \frac{4^4}{2}-\frac{1}{3(4)}-\frac{8(4)^{\frac32}}{3}+5(4) \right]-\left[ \frac12-\frac13-\frac83+5 \right] \\ &=\left[ 128-\frac{1}{12}-\frac{64}{3}+20 \right]-\left[ \frac12-\frac13-\frac83+5 \right] \\ &=\frac{1519}{12}-\frac52\\ &=\frac{1489}{12}\approx124.\overline{083} \end{aligned}$$

6. Given $$f(x)=\begin{cases}x^{3},&x >1\\ x+1,&x\leq 1\end{cases}$$, then find $$\int\limits_0^6 f(x)dx$$.

By using the formula that states the additivity with respect to endpoints as shown in Section, we can break the integral into two parts based on the conditions provided, for $$f(x)=x^3, ~ x>1 $$ and $$f(x)=x+1, ~ x\leq1 $$. Hence,$$\begin{aligned} \int\limits_0^6 f(x)dx&= \int\limits_0^1 x+1 dx+\int\limits_1^6 x^3dx \\ &= \left[\frac{x^2}{2}+x\right]^1_0+\left[ \frac{x^4}{4} \right]^6_1 \\ &=\left[ \frac{1^2}{2}+1 \right]+\left[ \frac{6^4}{4}-\frac{1^4}{4} \right] \\ &=\frac{3}{2}+\frac{1295}{4} \\ &=\frac{1301}{4}=325.25 \end{aligned}$$7. Let $$f(x)=\int\limits_1^xt^2dt$$. Then find $$f'(x)$$.

This given function can be found the derivative by using the Fundamental Theorem of Calculus Part I, as stated in Section .$$f'(x)=\frac{d}{dx}\left[\int\limits_1^x t^2 dt\right]=x^2.$$How is this $$x^2$$ coming from? This comes from the definite integral that we find using the second part of the Fundamental Theorem of Calculus. Since we know that the derivative is the opposite of what we do with the antiderivative, $$\begin{aligned} f'(x)=\frac{d}{dx}\left[\int\limits_1^x t^2 dt\right]&=\frac{d}{dx}\left[ \left.\frac{t^3}{3} \right|^x_1 \right] \\ &=\frac{d}{dx}\left[ \frac{x^3}{3} -\frac{1^3}{3}\right] \\ &=\frac{d}{dx}\left[ \frac{x^3}{3} \right]-\frac{d}{dx} \left[ \frac13 \right] \\ &= \frac{d}{dx}\left[ \frac{x^3}{3} \right]-0 \\ &=\frac13 \frac{d}{dx}[x^3] \\ &=\frac13\cdot3x^{3-1} \\ &=\frac1{\cancel3}\cdot\cancel{3}x^2 \\ &=x^2 \end{aligned}$$8. Given $$A(\theta)=\int\limits_1^{\theta^2}2x\cos(4x^2) dx$$. Then find $$A'(\theta)$$.

Here is another form of the function that is defined by the definite integral as a function. To find the derivative, we can apply the Chain Rule of the differentiation as in Section, which $$u=\theta^2$$, then $$\begin{aligned} A'(\theta)=\frac{d}{d\theta}\left[ \int\limits_1^{\theta^2}2x\cos(4x^2) dx \right] &= \frac{d}{d\theta}\left[ \int\limits_{1}^{u}2x\cos(4x^2) dx \right]\frac{du}{d\theta} \\ &=\left( 2u\cos(4u^2) \right)\frac{du}{d\theta} \\ &=\left( 2\theta^2\cos\left(4(\theta^2)^2\right) \right)\frac{d}{d\theta}\left[ \theta^2 \right] \\\ &=\left( 2\theta^2\cos(4\theta^4) \right)(2\theta) \\ &=4\theta^3\cos(4\theta) \end{aligned}$$9. If $$M(x)= \int\limits_{x^3}^x\cos^4(t)-\sin^2(t)dt$$. Then find $$M'(x)$$.

This is quite different from the two exercise problems previously. If previously we looked at the variable as being only at one bound of the integration, then what if the variable was placed on both the upper and lower bounds? By using the additive property of a definite integral, we can break up the integral into two parts and take the derivative to generates $$M'(x)$$ from given function.$$\frac{d}{dx}\left[ \int\limits_{x^3}^x\cos^4(t)-\sin^2(\theta)dt \right] =\frac{d}{dx}\left[ \int\limits_{x^3}^{0}\cos^4(t)-\sin^2(t)dt+\int\limits_0^{x}\cos^4(t)-\sin^2(t)dt \right]$$Therefore,$$\begin{aligned} \frac{d}{dx}\left[ \int\limits_{x^3}^{0}\cos^4(t)-\sin^2(t)dt+\int\limits_0^{x}\cos^4(t)-\sin^2(t)dt \right] &=\frac{d}{dx}\left[ \int\limits_{x^3}^{0}\cos^4(t)-\sin^2(t)dt \right]+\frac{d}{dx}\left[ \int\limits_0^{x}\cos^4(t)-\sin^2(t)dt \right] \\ &=-\frac{d}{dx}\left[ \int\limits_0^{x^3} \cos^4(t)-\sin^2(t)dt \right] +\frac{d}{dx} \left[ \int\limits_0^{x}\cos^4(t)-\sin^2(t)dt  \right] \\ &=-\left(\frac{d}{du} \left[ \int\limits_{0}^{u}\cos^4(t)-\sin^2(t)dt \right]\frac{d u}{dx}  \right) +\cos^4(x)-\sin^2(x) \\ &=-\left(\left( \cos^4(x^3)-\sin^2(x^3) \right)\frac{d}{dx}\left[ x^3 \right] \right) +\cos^4(x)-\sin^2(x) \\ &=-\left( \left[ \cos^4(x^3)-\sin(x^3) \right](3x^2) \right)+ \cos^4(x)-\sin^2(x) \\ &=-3x^2\cos^4(x^3)+3x^2\sin^2(x^3) + \cos^4(x)-\sin^2(x)  \end{aligned}$$10. For the function$${\displaystyle f(x)={\sqrt {x}}}$$ over the given closed interval, $${\displaystyle [4,9]}$$find the value(s) $$c$$ guaranteed by the mean value theorem for the definite integral.

To find the value(s)$$c$$ that are guaranteed by the mean value theorem for the definite integral, we can first find the $$f(c)$$ by using the formula shown below. $$f(c)=\frac{1}{b-a}\int\limits^b_a f(x) dx$$Such that,$$\begin{aligned}f(c)&=\frac{1}{9-4}\int\limits^9_4 \sqrt x dx \\ &=\frac15\left[ \frac{2x^{\frac{3}{2}}}{3} \right]^9_4 \\ &=\frac{1}{5}\left[ \frac{2(9)^{\frac{3}{2}}}{3}-\frac{2(4)^{\frac32}}{3} \right] \\ &=\frac15\left[ 18 -\frac{16}{3} \right] \\ &=\frac15\left[ \frac{38}{3} \right] \\\sqrt c &=\frac{38}{15} \end{aligned}$$Now, in the case of finding the value(s) c that are guaranteed by the mean value theorem for a definite integral, we can give the squared on both sides.$$\begin{aligned} \sqrt c &= \frac{38}{15} \\ \left( \sqrt{c} \right)^2 &= \left( \frac{38}{15} \right)^{2} \\ c &=\frac{1444}{225} \approx 6.\overline{417} \end{aligned}$$For the value of $$c$$ guaranteed by the Mean Value Theorem for Integrals is $$\frac{1444}{225} \approx 6.\overline{417}$$.